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Brilliant Sub Junior Calculus Contest (Season-1) Note-2

This is the second note to the season of Brilliant Sub Junior Calculus Contest (Season-1).

Click here for First Note (if you want to visit it)

Hi everyone! This is the first season of Brilliant Sub Junior Calculus Contest. This contest is for beginners or intermediate ones who want to sharpen their skills of problem-solving in overall calculus.

The aim of this contest is to improve the skill of in the computation in all sorts of problem (of basic level) in calculus like integrals (both definite and indefinite), differentiation, limits, ecetra by learning from each other and of course to have fun!

Eligibility:- People should fulfill either of the 2 following

  • 17 years or below

  • Level 4 or below in Calculus

    Eligible people here may participate in this contest.

The rules are as follows:

  • I will start by posting the first problem. If there is a user solves it, then they must post a new one.

  • You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

  • Only make substantial comment that will contribute to the discussion.

  • Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

  • If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

  • The scope of questions is only computation of basic level problems in calculus.

  • It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

  • You are also NOT allowed to post a solution using a contour integration or residue method.

Answer shouldn't contain any Special Function.

Please post your solution and your proposed problem in a single new thread.

Format your post is as follows:

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**SOLUTION OF PROBLEM xxx (number of problem) :**

**[Post your solution here]**

**PROBLEM xxx (number of problem) :**

**[Post your problem here]**

The comments will be easiest to follow if you sort by "Newest":

Note by Akshay Yadav
8 months, 1 week ago

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Problem 25

Prove that:

\[\dfrac{5050\displaystyle \int_0^1(1-x^{50})^{100}\,\mathrm{d}x}{\displaystyle \int_0^1(1-x^{50})^{101}\,\mathrm{d}x}=\boxed{5051}\]

This problem has been solved by Aditya Sharma. Rishabh Cool · 8 months, 1 week ago

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@Rishabh Cool Solution without using Beta function: \[\dfrac{5050\overbrace{\displaystyle \int_0^1(1-x^{50})^{100}\,\mathrm{d}x}^{\color{red}{I_1}}}{\underbrace{\displaystyle \int_0^1(1-x^{50})^{101}\,\mathrm{d}x}_{\color{red}{I_2}}}\]

Applying IBP on \(I_2\) \[I_2=x(1-x^{50})^{101}|_0^1+5050\displaystyle \int_0^1 x^{50}(1-x^{50})^{100}\,\mathrm{d}x\] \[=0+5050\displaystyle \int_0^1(\color{green}{1}-(\color{green}{1}- x^{50}))(1-x^{50})^{100}\,\mathrm{d}x\]

\[=5050\left(\displaystyle \int_0^1(1-x^{50})^{100}\,\mathrm{d}x-\displaystyle \int_0^1(1-x^{50})^{101}\,\mathrm{d}x\right)\]

\[\implies I_2=5050(I_1-I_2)\]

Rearranging we get:

\[\dfrac{5050I_1}{I_2}=\boxed{5051}\] Rishabh Cool · 8 months, 1 week ago

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@Rishabh Cool Both methods are same. The proof of Gamma functional equation uses IBP, so both methods are essentially the same. Samuel Jones · 8 months ago

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@Rishabh Cool Problem 25 : Let us first evaluate the numerator & denote it as \(I_1\)

Note : \(\mathbf{Beta}\) Function is used which is as follows just to recall :

\(\large \int_{0}^{1} x^{m-1}(1-x)^{n-1}=\beta(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\frac{(m-1)!(n-1)!}{(m+n-1)!}\) with m,n > 0.

Let \(1-x^{50}=t\)

\(-50x^{49}dx=dt\)

\(x^{49} = (1-t)^{\frac{49}{50}}\) & \(dx=\frac{dt}{-50(1-t)^{\frac{49}{50}}}\)

The integral is changed to :

\(I_1=-\frac{1}{50}\int_{1}^{0} t^{100}(1-t)^{-\frac{49}{50}}dt\)

\(50I_1=\int_{0}^{1} t^{101-1} (1-t)^{\frac{1}{50}-1}dt\)

\(50I_1=\beta(101,\frac{1}{50}=\frac{\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma(101+\frac{1}{50})}\)

Similarly the denominator just differs in case of the exponent of t, denote the denominator by \(I_2\) so we have ,

\(50I_2=\beta(102,\frac{1}{50})=\frac{\Gamma(102)\Gamma(\frac{1}{50})}{\Gamma(102+\frac{1}{50})}\)

\(50I_2=\frac{101\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma([1]+[101+\frac{1}{50}])}\)

Since we have \(\Gamma(n+1)=n\Gamma(n)\) so ,

\(50I_2= \frac{101}{101+\frac{1}{50}}\frac{\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma(101+\frac{1}{50})}\)

\(50I_2=\frac{101.50}{5051}.50I_1\)

\(5051=\frac{5050 . I_1}{I_2}\)

Therefore ,

\(\large \frac{5050.\int_{0}^{1}(1-x^{50})^{100}dx}{\int_{0}^{1}(1-x^{50})^{101}dx} = 5051\) \(\large \boxed{Proved}\) Aditya Sharma · 8 months, 1 week ago

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Problem 23 \[\large\int\dfrac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx=~?\]

This problem has been solved by Harsh Shrivastava. Rishabh Cool · 8 months, 1 week ago

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@Rishabh Cool

Harsh Shrivastava · 8 months, 1 week ago

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@Harsh Shrivastava A similar problem like this came in JEE Mains 2016! I just saw the question paper and recalled this problem ;) Nihar Mahajan · 8 months ago

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@Nihar Mahajan Yeah. Harsh Shrivastava · 8 months ago

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\(\large \mathbf{Problem}\) \(\large \mathfrak{27}\)

Prove that : \(\large \int_{0}^{\pi} \frac{xdx}{1+cos(\alpha)sin(x)} = \frac{\alpha\pi}{sin(\alpha)}\) Aditya Sharma · 8 months, 1 week ago

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@Aditya Sharma \[\large I=\int_{0}^{\pi} \frac{xdx}{1+\cos(\alpha)\sin(x)}\]

Using \(\int_a^b f(x)dx=\int_a^b f(a+b-x) dx\) and adding both integrals and simplifying gives:

\[I=\dfrac{\pi}{2}\left(\int_0^{\pi}\dfrac{dx}{1+(\cos \alpha)( \sin x)}\right)\] Using \(\sin x=\dfrac{2\tan^2 \frac x2}{1+\tan^2\frac x2}\) and substituting \(\tan^2 \frac x2=t\) such that \(\dfrac{(1+\tan^2 \frac x2 )dx}{2}=dt\) .

\[I={\pi}\int_{0}^{\infty}\left(\dfrac{dt}{1+t^2+2\cos \alpha t}\right)\]

\[I={\pi}\int_{0}^{\infty}\left(\dfrac{dt}{(t+\cos \alpha )^2+\sin^2 \alpha }\right)\]

Using \(\int \dfrac{dx}{x^2+a^2}=\frac 1a(\tan^{-1} \frac xa)\) .

\[=\dfrac{\pi}{\sin \alpha}\left(\tan^{-1}\left(\dfrac{t+\cos \alpha }{\sin \alpha}\right)\right)|_0^{\infty}\]

\[=\dfrac{\pi}{\sin \alpha}(\dfrac{\pi}{2}-\underbrace{\tan^{-1} \left(\cot \alpha\right)}_{\dfrac{\pi}{2}-\alpha})\]

\[\large \boxed{=\dfrac{\pi\alpha}{\sin \alpha}}\] Rishabh Cool · 8 months, 1 week ago

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Problem 30

\[\large\displaystyle \int \dfrac{\mathrm{d}x}{x^{11}\sqrt{1+x^4}}=\ ?\] Rishabh Cool · 8 months, 1 week ago

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@Rishabh Cool \( I = \displaystyle \int \frac{\mathrm{d}x}{x^{11}\sqrt{1+x^4}}= \)

Let \( \frac{1}{x^4} = t \)

So , \( I = - \frac{1}{4} \displaystyle \int \dfrac{t^2 \mathrm{d}t}{\sqrt{1+t}} \)

Adding and subtracting 1 in numerator , And integrating ,

We have \( I = - \frac{1}{4} {[ \frac{2}{15} (t+1)^{3/2} ( 3t -7) + 2\sqrt{t+1} ]} + C \)

After putting the value of t in terms of x and simplifying , Now , \( I = - \frac{ \sqrt{x^4 + 1}( 8x^8 - 4x^4 + 3)}{30x^{10}} + C \) Sachin Vishwakarma · 8 months, 1 week ago

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Problem 28 \[\large\mathfrak{S}=\sum_{n=1}^x\left(\dfrac{1}{n(n+1)(n+2)\cdots(n+p)}\right)\]

\[\dfrac{\displaystyle\int_0^{2016}\left(\left(\frac1{p^2(p-1)!}-\mathfrak S\right)\dfrac{p(x+p)!}{x!}\right)dx}{\displaystyle\int_0^{2}\left(\left(\frac1{p^2(p-1)!}-\mathfrak S\right)\dfrac{p(x+p)!}{(x-2)!}\right)dx}= \color{blue}{\varphi}\]

Find \(\dfrac{\sqrt{60^2-\color{blue}{\varphi}}}{3!}\).

**Original Rishabh Cool · 8 months, 1 week ago

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@Rishabh Cool Solution to Problem 28:

There is a quick technique to solve the summations of this kind..... Slash of the first term and write down the other terms as it is in deno. and multiply by the number of remaining terms in the denominator (here 'p') now multiply this result with (-1) and add a constant C so the result till now is : \(\\ S=C-\dfrac{1}{p(x+1)(x+2)\cdots(x+p)} \\ \)

When \(x=1, \ S=\dfrac{1}{(p+1)!} \ also \ S=C-\dfrac{1}{p(p+1)!} \\ \) This gives \(C=\dfrac{1}{p.p!}\)

In both the integrals \(\dfrac{1}{p^2(p-1)!}-S\) is equivalent to \(C-(C-\dfrac{1}{p(x+1)(x+2)\cdots(x+p)}) \\ \) which is equivalent to \(\dfrac{x!}{p(x+p)!}\)

Now the integral in numerator turns into

\(\displaystyle \int_0^{2016} dx\) and that in denominator turns into

\(\displaystyle \int_0^{2} (x)(x-1)dx\)

Thus, \(\varphi=3024 \) and answer is 4. Samarth Agarwal · 8 months, 1 week ago

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\(\mathbf{Problem}\) \(\mathbf{26}\)

Let \(f(x)\) be a function satisfying \(f(0)=2,f'(0)=3,f''(x)=f(x)\) , then prove that :

\(f(4) = \frac{5(e^8-1)}{2e^4}\) Aditya Sharma · 8 months, 1 week ago

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@Aditya Sharma Let us consider a polynomial \( f(x) = a + bx + cx^2 + dx^3 + ex^4 + fx^5 + gx^6 + ....... \).

\( f'(x) = b + 2cx + 3dx^2 + 4ex^3 + 5fx^4 +6gx^5 + 7hx^6 + ....... \).

\( f''(x) = 2c + 6dx + 12ex^2 + 20fx^3 + 30gx^4 + 42hx^5 + ....... \).

Since , \( f(0) = 2 \implies a=2 \) and \( f'(0) = 3 \implies b=3 \).

On comparing coefficient of \( x^n \) in equation \( f(x) = f''(x) \)

\( c=1 , d= \frac{1}{2} , e =\frac{1}{12} , f=\frac{1}{40} , g=\frac{1}{360} , h =\frac{1}{1680} \)

We have , \( f(4) = 2 + 12 + 16 +\frac{4^3}{2}+\frac{4^4}{12} +\frac{4^5}{40} + \frac{4^6}{360} \).

Or \( f(4) = \displaystyle \sum_{n=1}^\infty \frac{ 4^n ( (-1)^n + 5 ) }{ 8(n-1)!} = \displaystyle \sum_{n=1}^\infty \frac{ 4^n (-1)^n }{ 8(n-1)!} + \frac{5}{2} \displaystyle \sum_{n=1}^\infty \frac{ 4^{(n-1)} }{ (n-1)!} = \frac{1}{8} \frac{-4}{e^4} + \frac{5}{2} e^4 = \boxed{ \frac{ 5e^8 - 1}{2e^4} } \) Sachin Vishwakarma · 8 months, 1 week ago

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@Aditya Sharma The problem is incorrect. If a polynomial \(f(x)\) satisfies \(f''(x) = f(x)\), then, even without solving the differential equation, the number of roots of \(f(x)\) equals the number of roots of \(f''(x)\), which is a contradiction. So it must be stated that \(f\) is a function, not a polynomial function. Samuel Jones · 8 months ago

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@Samuel Jones This is correct ^. In reality the function is exponential. Vighnesh Shenoy · 8 months ago

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@Aditya Sharma The problem is of a book named "Integral calculus by Amit M Agaarwal" The solution he gave I am just posting it here.Now I am finding the mistake in his or sachin's solution.

\(f''(x)=f(x)\implies 2f'(x)f''(x)=2f(x)f'(x)\implies d({f'(x)}^2) = d({f(x)}^2)\)

\(\implies {f'(x)}^2 = {f(x)}^2 + C\)

Utilising the conditions for \(f(0) = 2 , f'(0) = 3\) we get \(C=5\)

We have \(f'(x) = \sqrt{5+{f(x)}^2}\)

\(\int dx = \int \frac{d({f(x)})}{\sqrt{5+{f(x)}^2}}\)

\(x + C_1 = log|f(x) + \sqrt{5+{f(x)}^2}|\)

again by the conditions putting x=0 we get ,

\(C_1=log5\)

So, \(x=log\frac{|f(x) + \sqrt{5+{f(x)}^2}|}{5}\) ..................................... (1)

Observe that , \(log\frac{|f(x) + \sqrt{5+{f(x)}^2}|}{5} = log\frac{5}{|\sqrt{5+{f(x)}^2}-f(x)|}\)

From (1) ,

\(5e^x = f(x) + \sqrt{5+{f(x)}^2}\) & similarly \(5e^{-x}=\sqrt{5+{f(x)}^2}-f(x)\)

Extracting f(x) we get ,

\(2f(x) = 5(e^x-e^{-x})\implies f(x) = \frac{5(e^x-e^{-x})}{2}\)

\(\large\boxed{ f(4) = \frac{5(e^4-e^{-4})}{2} = \frac{5(e^8-1)}{2e^4}}\) Aditya Sharma · 8 months, 1 week ago

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Problem 24:

Evaluate \[\displaystyle \int \dfrac{dx}{(\ln^{3} (\tan x))(\sin 2x)} \]

This problem was solved by Rishabh Cool. Harsh Shrivastava · 8 months, 1 week ago

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@Harsh Shrivastava Solution to Problem 24

Substitute \(\ln \tan x=t\) such that \(\dfrac{dx}{\sin 2x}=\dfrac{dt}{2}\). Integral transforms to: \[\int \dfrac{dt}{2 t^3}\] \[=\dfrac{-1}{ 4t^2}=\dfrac{-1}{4\ln^2 \tan x}+C\] Rishabh Cool · 8 months, 1 week ago

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Problem 31

If \( I_{1} = \displaystyle \int_0^{\frac{\pi}{2} } f( \sin 2x) \sin x \mathrm{d}x \) and \( I_{2} = \displaystyle \int_0^{\frac{\pi}{4} } f( \cos 2x) \cos x \mathrm{d}x \) , then find \( \frac{I_{1}}{I_{2}}\) ? Sachin Vishwakarma · 8 months, 1 week ago

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@Sachin Vishwakarma \( I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{2}} f(\sin 2x)\sin x dx \)
Using the property,
\( \displaystyle \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)+f(2a-x)dx \)
\( I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x ) dx \)
In \( I_{2} \) ,
Use \( \displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b} f(a+b-x)dx \)
\( \therefore I_{2} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)\cos\left( \dfrac{\pi}{4} - x\right) dx \)
\( \therefore I_{2} = \dfrac{1}{\sqrt{2}}\cdot \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x) dx = \dfrac{I_{1}}{\sqrt{2}} \)
\( \dfrac{I_{1}}{I_{2}} = \sqrt{2} \) Vighnesh Shenoy · 8 months, 1 week ago

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Problem 29:

Prove that:

\[\displaystyle \lim_{x\to 0} \dfrac{24}{x^3} \displaystyle \int_0^x \dfrac{t \ln(1+t)}{t^4+4} dt =2 \] Samarth Agarwal · 8 months, 1 week ago

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@Samarth Agarwal Applying LHopital:

\[\lim_{x\to 0}\dfrac{8x\ln(1+x)}{x^2(x^4+4)}\]

\[=8\lim_{x\to 0}\dfrac{\ln(1+x)}{x(4)}\]

\[\text{Using} \lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1\]

\[=\boxed 2\] Rishabh Cool · 8 months, 1 week ago

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\(\text{Problem 38}\)\[\]What is the value of the integral given below?\[\int\dfrac{\tan x}{a+b\tan^2x}dx\] Adarsh Kumar · 8 months ago

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@Adarsh Kumar My method was to take denominator as t and then apply a little partial fraction giving the same answer... :)

@Aditya Sharma you may post the next question Samarth Agarwal · 8 months ago

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@Adarsh Kumar \(\large I = \int \frac{tanx}{a+b(tan)^2x}\ = \int \frac{sin(x)cos(x)}{acos^2x+bsin^2x}\)

\(\text{Put } sin^2x=t\)

\(I = \frac{1}{2}\int \frac{sin(2x)}{a - sin^2x(a-b)} = \int \frac{dt}{2(a - t(a-b))} = \frac{1}{2(b-a)}ln|a-t(a-b)| + C\) Aditya Sharma · 8 months ago

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Problem 37

Let \(f(x) \) be a differentiable function such that \( \displaystyle \dfrac{d f(x) }{dx} = f(x) + \int_0^2 f(x) \, dx \) and \(f(0) = \dfrac{4-e^2}3 \), then prove: \[f(2)=\dfrac{2e^2+1}{3}\] Rishabh Cool · 8 months ago

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@Rishabh Cool

Samarth Agarwal · 8 months ago

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@Rishabh Cool I don't have a good problem right now.... Can anybody else post the question... Thanx Samarth Agarwal · 8 months ago

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@Samarth Agarwal Can i post the next question? Adarsh Kumar · 8 months ago

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@Adarsh Kumar Yeah! Anybody is welcome to do so. Akshay Yadav · 8 months ago

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@Adarsh Kumar \(\large I = \int \frac{tanx}{a+b(tan)^2x}\ = \int \frac{sin(x)cos(x)}{acos^2x+bsin^2x}\)

\(\text{Put } sin^2x=t\)

\(I = \frac{1}{2}\int \frac{sin(2x)}{a - sin^2x(a-b)} = \int \frac{dt}{2(a - t(a-b))} = \frac{1}{2(b-a)}ln|a-t(a-b)| + C\)

Like this ? Aditya Sharma · 8 months ago

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@Aditya Sharma Just cut and paste it to Adarsh's new comment..:-) Rishabh Cool · 8 months ago

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@Rishabh Cool Yup ! Aditya Sharma · 8 months ago

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@Aditya Sharma @Adarsh Kumar under the heading comments your name is written with a box below it.... U have to write in it Samarth Agarwal · 8 months ago

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@Samarth Agarwal Is this fine?(i have posted) Adarsh Kumar · 8 months ago

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@Rishabh Cool Yup!Just put that expression in mod,and add the integration constant :) Adarsh Kumar · 8 months ago

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@Adarsh Kumar @Adarsh Kumar pls post it separately... Thanks Samarth Agarwal · 8 months ago

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@Samarth Agarwal Separately meaning,not in response to anybody? Adarsh Kumar · 8 months ago

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@Adarsh Kumar Pls post it separately... Thanks Samarth Agarwal · 8 months ago

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@Akshay Yadav Can we change the note after 40 problems as my phone has already started lagging Samarth Agarwal · 8 months ago

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@Samarth Agarwal My tablet too is going crazy. I will change it tomorrow (even if the question number doesn't crosses 40). Till then let's hope it the question number crosses 40! Akshay Yadav · 8 months ago

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@Samarth Agarwal Same is happening with my phone too... Rishabh Cool · 8 months ago

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@Adarsh Kumar Yes if @Akshay Yadav allows Samarth Agarwal · 8 months ago

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@Samarth Agarwal Sometimes I don't get differential equations, what should I do? Akshay Yadav · 8 months ago

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@Akshay Yadav Only practice helps in this topic..... And there are some standard rules which are needed to be followed Samarth Agarwal · 8 months ago

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@Rishabh Cool @Rishabh Cool is this your original problem??? Btw Its really very nice Samarth Agarwal · 8 months ago

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@Samarth Agarwal Nope... :-).... I'm not sure but it might be an IIT JEE problem.... I found this problem interesting so I found it worth sharing. Rishabh Cool · 8 months ago

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PROBLEM 33 \[\displaystyle \int \sqrt{\tan x} \ dx\]

The answer is very long so be patient.! Samarth Agarwal · 8 months ago

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@Samarth Agarwal \( \tan x = t^{2} \)
\( dx = \dfrac{2t}{1+t^{4}}dt \)
\( I = \displaystyle \int \dfrac{2t^{2}}{1+t^{4}}dt \)
\( I = \displaystyle \int \dfrac{t^{2}-1}{1+t^{4}}dt + \int \dfrac{t^{2}+1}{1+t^{4}}dt \)

\( I = \displaystyle \int \dfrac{1-\frac{1}{t^{2}}}{t^{2} + \dfrac{1}{t^{2}}}dt + \int \dfrac{1+\frac{1}{t^{2}}}{t^{2} + \dfrac{1}{t^{2}}}dt \)

\( I = \displaystyle \int \dfrac{1-\frac{1}{t^{2}}}{\left(t+\frac{1}{t} \right)^{2} - 2}dt + \int \dfrac{1+\frac{1}{t^{2}}}{\left(t-\dfrac{1}{t}\right)^{2} + 2 }dt \)
\( I = \displaystyle \int \dfrac{du}{u^{2} - (\sqrt{2})^{2}} + \int \dfrac{dv}{v^{2}+(\sqrt{2})^{2}} \)
\( I =\dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{u-\sqrt{2}}{u+\sqrt{2}}\right) + \dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{v}{\sqrt{2}}\right) \)

\( I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)+ \dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{t-\frac{1}{t}}{\sqrt{2}}\right) \)

\( I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{\sqrt{\tan(x)} + \sqrt{\cot(x)}-\sqrt{2}}{\sqrt{\tan(x)} + \sqrt{\cot(x)}+\sqrt{2}}\right)+\dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{\sqrt{\tan(x)}-\sqrt{\cot(x)}}{\sqrt{2}}\right) + c \) Vighnesh Shenoy · 8 months ago

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@Vighnesh Shenoy Will you post next problem or shall we post? You often stall the contest.

(Sorry to be rude) Harsh Shrivastava · 8 months ago

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@Harsh Shrivastava You post. I have been trying to come up with a good problem, but I haven't been getting one. Vighnesh Shenoy · 8 months ago

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@Vighnesh Shenoy Okay according to rules @Samarth Agarwal should post the next problem. Harsh Shrivastava · 8 months ago

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@Vighnesh Shenoy Very nice..... Pls post the next question Samarth Agarwal · 8 months ago

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Let 's define the term,
\( I_{n} = \displaystyle \int_{0}^{\infty} \dfrac{\arctan\left(\frac{2x}{1}\right) + \arctan\left(\frac{3x}{2}\right) + \ldots \arctan\left(\frac{(n+1)x}{n}\right) - n\arctan(x)}{x} dx \) for integer n.

Find \( I_{20} - I_{6} \) Vighnesh Shenoy · 8 months, 1 week ago

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@Samuel Jones The answer is not 0.
\( I_{n} = \displaystyle \int_{0}^{\infty} \dfrac{ \displaystyle \sum_{k=1}^{n} \left(\arctan\left(\dfrac{(k+1)x}{k}\right) -\arctan(x) \right)}{x}dx \)

Consider the integral,
\( J(a) = \displaystyle \int_{0}^{\infty} \dfrac{ \arctan(ax) - \arctan(x) }{x} dx \)
\( \dfrac{dJ}{da} = \displaystyle \int_{0}^{\infty} \dfrac{\partial}{\partial a} \dfrac{\arctan(ax)-\arctan(x)}{x} dx \)
\( \dfrac{dJ}{da} = \displaystyle \int_{0}^{\infty} \dfrac{1}{1+(ax)^{2}}dx \)
\( \dfrac{dJ}{da} = \dfrac{\pi}{2a} \)
\( \therefore J = \dfrac{\pi \ln(a)}{2} + c \)
Since, \( J(1)= 0 , \rightarrow c = 0 \)
\( \therefore J(a) = \dfrac{\pi \ln(a)}{2} \)
\( I_{n} = \displaystyle \sum_{k=1}^{n} J\left(\dfrac{k+1}{k}\right) = \dfrac{\pi \ln(n+1)}{2} \)
\( I_{20} - I_{6} = \dfrac{\pi}{2} \left( \ln(21) - \ln(7) \right) = \dfrac{\pi \ln(3)}{2} \)


I can differentiate under the integral when the function is continuous, and differentiable. As for why your answer is incorrect,
After substituting \( x = \dfrac{1}{t} \)
\( \arctan\left(\dfrac{(k+1)x}{k}\right) = \arctan\left(\dfrac{(k+1)}{kt}\right) = \dfrac{\pi}{2} - \arctan\left(\dfrac{kt}{(k+1)}\right) \ne \dfrac{\pi}{2} - \arctan\left(\dfrac{(k+1)x}{k}\right) \) Vighnesh Shenoy · 8 months ago

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\(\mathbf{Problem} \mathfrak{39}\) Determine : \(\int \frac{dx}{secx+2sinx}\) Aditya Sharma · 8 months ago

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@Aditya Sharma The given integral can be written as

\(\displaystyle \int \dfrac{\cos x \ dx}{1+2 \sin x \cos x}\)

\(\displaystyle = \int \dfrac{\cos x \ dx}{(\sin x + \cos x )^2}\)

\(\displaystyle = \dfrac{1}{2} \int \dfrac{(\sin x + \cos x) + (\cos x - \sin x)}{(\sin x + \cos x )^2} dx\)

\( \displaystyle = \dfrac{1}{2} \int \dfrac{dx}{\sin x + \cos x} + \dfrac{1}{2} \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2}\)

\( \displaystyle = \dfrac{1}{2\sqrt{2}} \int \csc \left(x + \frac{\pi}{4}\right) dx + \dfrac{1}{2} \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2}\)

\( \displaystyle = -\dfrac{1}{2\sqrt{2}} \ln \left| \csc\left( x +\frac{\pi}{4}\right) + \cot \left(x+\frac{\pi}{4}\right) \right|-\dfrac{1}{2(\sin x +\cos x)} + C\) Samuel Jones · 8 months ago

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@Samuel Jones My method was a little bit different!Although nice work! Adarsh Kumar · 8 months ago

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@Samuel Jones Yeah same form as @Adarsh Kumar , He just simplified the last one to \(\frac{1}{2\sqrt2}\frac{1}{sin(x+\frac{\pi}{4})}\) Aditya Sharma · 8 months ago

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@Aditya Sharma I think @Akshay Yadav now the note should be changed and the next note will begin with problem 40 as my PC,tablet,Mobile everything is at a standstill :P Aditya Sharma · 8 months ago

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@Aditya Sharma Is the answer long? Adarsh Kumar · 8 months ago

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@Adarsh Kumar Not that really. Not as that of \(\sqrt{tanx}\) Aditya Sharma · 8 months ago

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@Aditya Sharma Yeah not that long but does it contain log and some trigo functions? Adarsh Kumar · 8 months ago

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@Adarsh Kumar Yeah it has log & trigo functions Aditya Sharma · 8 months ago

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@Aditya Sharma Is it \[\dfrac{1}{2\sqrt{2}}\left(\ln |\csc(x+\frac{\pi}{4})-\cot(x+\frac{\pi}{4})|-\dfrac{1}{\sin(x+\frac{\pi}{4})}\right)+C\] Adarsh Kumar · 8 months ago

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@Adarsh Kumar Absolutely ! Great work. Aditya Sharma · 8 months ago

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@Aditya Sharma Haha thanx!It was a good question!I am posting the solution! Adarsh Kumar · 8 months ago

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@Adarsh Kumar Yeah ! Aditya Sharma · 8 months ago

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@Aditya Sharma @Aditya Sharma @Akshay Yadav Who will post the next Problem? Me or @Adarsh Kumar He had posted the answer while I was posting the solution, but the solution was posted by me earlier. Samuel Jones · 8 months ago

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@Samuel Jones I am a bit confused here, Accordingly @Samuel Jones should be allowed as he provided the solution a bit earlier. Aditya Sharma · 8 months ago

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@Aditya Sharma Okay. Let @Akshay Yadav post the next note. I'll post Problem \(40\) there. Samuel Jones · 8 months ago

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@Samuel Jones Yeah thanx :P or else my phone would have exploded . @Akshay Yadav please post the next note :) Aditya Sharma · 8 months ago

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PROBLEM 36

Let \(f(x)\) be a differentiable function and \(f'(x)=-f(x) , f'(x) =g(x) \)

Let \(h(x)=[f(x)]^2+[g(x)]^2 \) and h(5)=11 find h(10) Samarth Agarwal · 8 months ago

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@Samarth Agarwal \(\dfrac{f'(x)}{f(x)}=-1\)

Integrate both sides to get: \[f(x)=Ce^{-x}\] \[\implies g(x)=f'(x)=-Ce^{-x}\] \[\implies h(x)=2C^2e^{-2x}\]

\[\dfrac{h(10)}{h(5)}=\dfrac{e^{-20}}{e^{-10}}=e^{-10}\]

\[\implies h(10)=\boxed{11e^{-10}}\]

Is it or not??? Rishabh Cool · 8 months ago

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@Rishabh Cool Absolutely correct!!! You may post the next question :) Samarth Agarwal · 8 months ago

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PROBLEM 35:

Time for an easy problem!

Find the radius and coordinates of center of curvature of the curve

\[y=(4x-x^2-3)^{\frac{1}{2}}\]

at the points \(x=1.2\) and \(x=2\). Also what is this curve? Akshay Yadav · 8 months ago

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@Akshay Yadav It is a semi circle above x axis as y is always positive with radius 1 unit and center (2,0) Samarth Agarwal · 8 months ago

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@Samarth Agarwal Wow, I had expected a fast reply but you did it extremely fast! Akshay Yadav · 8 months ago

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@Akshay Yadav This is because it just requires simple co ordinate geometry Samarth Agarwal · 8 months ago

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@Samarth Agarwal I think that question is more based on co ordinate geometry as curve is just a semi circle Samarth Agarwal · 8 months ago

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PROBLEM 34:

Find the minimum value of the integral \(\displaystyle \int_0^{\pi} \left(x - \pi a - \frac {b}{\pi}\cos x\right)^2 \ dx \) Nihar Mahajan · 8 months ago

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@Nihar Mahajan I had simplified the expression and then integrated it, after simplification you get-

\(\displaystyle \int_0^{\pi} \left(x^2+\pi^2a^2+\frac{b^2}{\pi^2}\cos^2 (x)-2\pi ax+2ab\cos (x)-\frac{2b}{\pi}\cos (x) \right) dx\)

From that you get the expression-

\(\frac{3b^2+24b+6{\pi}^4a^2-6{\pi}^4a+2{\pi}^4}{6{\pi}}\)

Let this equals to \(y\) and then we use partial differentiation,

From that we get \(a=\frac{1}{2}\) and \(b=-4\).

Place the values back and then you'll get the required minimum value of the integral as,

\[\boxed{\frac{\pi^3}{12}-\frac{8}{\pi}}\] Akshay Yadav · 8 months ago

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Problem no : 32

\( f(x) = 3e^{x} + \dfrac{1}{1+x^{2}} \)
Let \( g(x) \) be the inverse of \( f(x) \).
Compute,
\( \displaystyle \int_{3}^{3e - \frac{1}{2}}g(x+1)dx \) Vighnesh Shenoy · 8 months, 1 week ago

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@Vighnesh Shenoy \(I = \displaystyle \int_{3}^{3e - \frac{1}{2}} g(x+1) dx = \int_{4}^{3e+\frac{1}{2}} g(x)dx = \int_{f(0)}^{f(1)}g(x)dx \)
Substitute, \( g(x) = t \rightarrow f(g(x)) = f(t) \)
But since, \( f(x) \) and \( g(x) \) are inverses of each other, \( f(g(x))= x \)
\( \therefore x = f(t) \rightarrow dx = f'(t)dt \)
\( \therefore I = \displaystyle \int_{0}^{1} tf'(t)dt \)
Integrating by parts,
\( I = \left[tf(t)\right]_{0}^{1} - \displaystyle \int_{0}^{1} f(t)dt \)
\( I = 3e + \dfrac{1}{2} - \displaystyle \int_{0}^{1} 3e^{t} + \dfrac{1}{1+t^{2}}dt = 3e + \dfrac{1}{2} - \left[ 3e^{t} + \tan^{1}(x) \right]_{0}^{1} \)
\( \therefore I = 3e + \dfrac{1}{2} -3e - \dfrac{\pi}{4} + 3 + 0 = \dfrac{7}{2} - \dfrac{\pi}{4} = \dfrac{14-\pi}{4} \) Vighnesh Shenoy · 8 months, 1 week ago

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Third note of this season has been posted. Click here to access it. Akshay Yadav · 8 months ago

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@Akshay Yadav @Samuel Jones Problem 40 on the new note please Aditya Sharma · 8 months ago

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