This is the second note to the season of Brilliant Sub Junior Calculus Contest (Season1).
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 problemsolving 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:
1 2 3 4 5 6 7 

The comments will be easiest to follow if you sort by "Newest":
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Top NewestProblem 25
Prove that:
\[\dfrac{5050\displaystyle \int_0^1(1x^{50})^{100}\,\mathrm{d}x}{\displaystyle \int_0^1(1x^{50})^{101}\,\mathrm{d}x}=\boxed{5051}\]
This problem has been solved by Aditya Sharma. – Rishabh Cool · 8 months, 1 week ago
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Applying IBP on \(I_2\) \[I_2=x(1x^{50})^{101}_0^1+5050\displaystyle \int_0^1 x^{50}(1x^{50})^{100}\,\mathrm{d}x\] \[=0+5050\displaystyle \int_0^1(\color{green}{1}(\color{green}{1} x^{50}))(1x^{50})^{100}\,\mathrm{d}x\]
\[=5050\left(\displaystyle \int_0^1(1x^{50})^{100}\,\mathrm{d}x\displaystyle \int_0^1(1x^{50})^{101}\,\mathrm{d}x\right)\]
\[\implies I_2=5050(I_1I_2)\]
Rearranging we get:
\[\dfrac{5050I_1}{I_2}=\boxed{5051}\] – Rishabh Cool · 8 months, 1 week ago
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– Samuel Jones · 8 months ago
Both methods are same. The proof of Gamma functional equation uses IBP, so both methods are essentially the same.Log in to reply
Note : \(\mathbf{Beta}\) Function is used which is as follows just to recall :
\(\large \int_{0}^{1} x^{m1}(1x)^{n1}=\beta(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\frac{(m1)!(n1)!}{(m+n1)!}\) with m,n > 0.
Let \(1x^{50}=t\)
\(50x^{49}dx=dt\)
\(x^{49} = (1t)^{\frac{49}{50}}\) & \(dx=\frac{dt}{50(1t)^{\frac{49}{50}}}\)
The integral is changed to :
\(I_1=\frac{1}{50}\int_{1}^{0} t^{100}(1t)^{\frac{49}{50}}dt\)
\(50I_1=\int_{0}^{1} t^{1011} (1t)^{\frac{1}{50}1}dt\)
\(50I_1=\beta(101,\frac{1}{50}=\frac{\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma(101+\frac{1}{50})}\)
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 ,
\(\large \frac{5050.\int_{0}^{1}(1x^{50})^{100}dx}{\int_{0}^{1}(1x^{50})^{101}dx} = 5051\) \(\large \boxed{Proved}\) – Aditya Sharma · 8 months, 1 week ago
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Problem 23 \[\large\int\dfrac{x^21}{x^3\sqrt{2x^42x^2+1}}\,dx=~?\]
This problem has been solved by Harsh Shrivastava. – Rishabh Cool · 8 months, 1 week ago
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– Nihar Mahajan · 8 months ago
A similar problem like this came in JEE Mains 2016! I just saw the question paper and recalled this problem ;)Log in to reply
– Harsh Shrivastava · 8 months ago
Yeah.Log in to reply
\(\large \mathbf{Problem}\) \(\large \mathfrak{27}\)
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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Using \(\int_a^b f(x)dx=\int_a^b f(a+bx) dx\) 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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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 \[\large\mathfrak{S}=\sum_{n=1}^x\left(\dfrac{1}{n(n+1)(n+2)\cdots(n+p)}\right)\]
\[\dfrac{\displaystyle\int_0^{2016}\left(\left(\frac1{p^2(p1)!}\mathfrak S\right)\dfrac{p(x+p)!}{x!}\right)dx}{\displaystyle\int_0^{2}\left(\left(\frac1{p^2(p1)!}\mathfrak S\right)\dfrac{p(x+p)!}{(x2)!}\right)dx}= \color{blue}{\varphi}\]
Find \(\dfrac{\sqrt{60^2\color{blue}{\varphi}}}{3!}\).
**Original – Rishabh Cool · 8 months, 1 week ago
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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 \(\dfrac{1}{p^2(p1)!}S\) 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
\(\displaystyle \int_0^{2} (x)(x1)dx\)
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 :
\(f(4) = \frac{5(e^81)}{2e^4}\) – Aditya Sharma · 8 months, 1 week ago
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\( f'(x) = b + 2cx + 3dx^2 + 4ex^3 + 5fx^4 +6gx^5 + 7hx^6 + ....... \).
\( f''(x) = 2c + 6dx + 12ex^2 + 20fx^3 + 30gx^4 + 42hx^5 + ....... \).
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 \( f(4) = \displaystyle \sum_{n=1}^\infty \frac{ 4^n ( (1)^n + 5 ) }{ 8(n1)!} = \displaystyle \sum_{n=1}^\infty \frac{ 4^n (1)^n }{ 8(n1)!} + \frac{5}{2} \displaystyle \sum_{n=1}^\infty \frac{ 4^{(n1)} }{ (n1)!} = \frac{1}{8} \frac{4}{e^4} + \frac{5}{2} e^4 = \boxed{ \frac{ 5e^8  1}{2e^4} } \) – Sachin Vishwakarma · 8 months, 1 week ago
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– Samuel Jones · 8 months ago
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.Log in to reply
– Vighnesh Shenoy · 8 months ago
This is correct ^. In reality the function is exponential.Log in to reply
\(f''(x)=f(x)\implies 2f'(x)f''(x)=2f(x)f'(x)\implies d({f'(x)}^2) = d({f(x)}^2)\)
\(\implies {f'(x)}^2 = {f(x)}^2 + C\)
Utilising the conditions for \(f(0) = 2 , f'(0) = 3\) we get \(C=5\)
We have \(f'(x) = \sqrt{5+{f(x)}^2}\)
\(\int dx = \int \frac{d({f(x)})}{\sqrt{5+{f(x)}^2}}\)
\(x + C_1 = logf(x) + \sqrt{5+{f(x)}^2}\)
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 ,
\(2f(x) = 5(e^xe^{x})\implies f(x) = \frac{5(e^xe^{x})}{2}\)
\(\large\boxed{ f(4) = \frac{5(e^4e^{4})}{2} = \frac{5(e^81)}{2e^4}}\) – 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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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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Using the property,
\( \displaystyle \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)+f(2ax)dx \)
\( I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x ) dx \)
In \( I_{2} \) ,
Use \( \displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b} f(a+bx)dx \)
\( \therefore I_{2} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)\cos\left( \dfrac{\pi}{4}  x\right) dx \)
\( \therefore I_{2} = \dfrac{1}{\sqrt{2}}\cdot \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x) dx = \dfrac{I_{1}}{\sqrt{2}} \)
\( \dfrac{I_{1}}{I_{2}} = \sqrt{2} \) – 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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\[\lim_{x\to 0}\dfrac{8x\ln(1+x)}{x^2(x^4+4)}\]
\[=8\lim_{x\to 0}\dfrac{\ln(1+x)}{x(4)}\]
\[\text{Using} \lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1\]
\[=\boxed 2\] – 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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@Aditya Sharma you may post the next question – Samarth Agarwal · 8 months ago
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\(\text{Put } sin^2x=t\)
\(I = \frac{1}{2}\int \frac{sin(2x)}{a  sin^2x(ab)} = \int \frac{dt}{2(a  t(ab))} = \frac{1}{2(ba)}lnat(ab) + C\) – 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 \(f(0) = \dfrac{4e^2}3 \), then prove: \[f(2)=\dfrac{2e^2+1}{3}\] – Rishabh Cool · 8 months ago
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– Samarth Agarwal · 8 months ago
I don't have a good problem right now.... Can anybody else post the question... ThanxLog in to reply
– Adarsh Kumar · 8 months ago
Can i post the next question?Log in to reply
– Akshay Yadav · 8 months ago
Yeah! Anybody is welcome to do so.Log in to reply
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\(\text{Put } sin^2x=t\)
\(I = \frac{1}{2}\int \frac{sin(2x)}{a  sin^2x(ab)} = \int \frac{dt}{2(a  t(ab))} = \frac{1}{2(ba)}lnat(ab) + C\)
Like this ? – Aditya Sharma · 8 months ago
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– Rishabh Cool · 8 months ago
Just cut and paste it to Adarsh's new comment..:)Log in to reply
– Aditya Sharma · 8 months ago
Yup !Log in to reply
– Samarth Agarwal · 8 months ago
@Adarsh Kumar under the heading comments your name is written with a box below it.... U have to write in itLog in to reply
– Adarsh Kumar · 8 months ago
Is this fine?(i have posted)Log in to reply
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– Adarsh Kumar · 8 months ago
Yup!Just put that expression in mod,and add the integration constant :)Log in to reply
@Adarsh Kumar pls post it separately... Thanks – Samarth Agarwal · 8 months ago
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– Adarsh Kumar · 8 months ago
Separately meaning,not in response to anybody?Log in to reply
– Samarth Agarwal · 8 months ago
Pls post it separately... ThanksLog in to reply
– Samarth Agarwal · 8 months ago
Can we change the note after 40 problems as my phone has already started laggingLog in to reply
– Akshay Yadav · 8 months ago
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!Log in to reply
– Rishabh Cool · 8 months ago
Same is happening with my phone too...Log in to reply
@Akshay Yadav allows – Samarth Agarwal · 8 months ago
Yes ifLog in to reply
– Akshay Yadav · 8 months ago
Sometimes I don't get differential equations, what should I do?Log in to reply
– Samarth Agarwal · 8 months ago
Only practice helps in this topic..... And there are some standard rules which are needed to be followedLog in to reply
@Rishabh Cool is this your original problem??? Btw Its really very nice – Samarth Agarwal · 8 months ago
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– Rishabh Cool · 8 months ago
Nope... :).... I'm not sure but it might be an IIT JEE problem.... I found this problem interesting so I found it worth sharing.Log in to reply
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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\( dx = \dfrac{2t}{1+t^{4}}dt \)
\( I = \displaystyle \int \dfrac{2t^{2}}{1+t^{4}}dt \)
\( I = \displaystyle \int \dfrac{t^{2}1}{1+t^{4}}dt + \int \dfrac{t^{2}+1}{1+t^{4}}dt \)
\( I = \displaystyle \int \dfrac{1\frac{1}{t^{2}}}{t^{2} + \dfrac{1}{t^{2}}}dt + \int \dfrac{1+\frac{1}{t^{2}}}{t^{2} + \dfrac{1}{t^{2}}}dt \)
\( I = \displaystyle \int \dfrac{1\frac{1}{t^{2}}}{\left(t+\frac{1}{t} \right)^{2}  2}dt + \int \dfrac{1+\frac{1}{t^{2}}}{\left(t\dfrac{1}{t}\right)^{2} + 2 }dt \)
\( I = \displaystyle \int \dfrac{du}{u^{2}  (\sqrt{2})^{2}} + \int \dfrac{dv}{v^{2}+(\sqrt{2})^{2}} \)
\( I =\dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{u\sqrt{2}}{u+\sqrt{2}}\right) + \dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{v}{\sqrt{2}}\right) \)
\( I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{t+\frac{1}{t}\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)+ \dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{t\frac{1}{t}}{\sqrt{2}}\right) \)
\( I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{\sqrt{\tan(x)} + \sqrt{\cot(x)}\sqrt{2}}{\sqrt{\tan(x)} + \sqrt{\cot(x)}+\sqrt{2}}\right)+\dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{\sqrt{\tan(x)}\sqrt{\cot(x)}}{\sqrt{2}}\right) + c \) – Vighnesh Shenoy · 8 months ago
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(Sorry to be rude) – Harsh Shrivastava · 8 months ago
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– Vighnesh Shenoy · 8 months ago
You post. I have been trying to come up with a good problem, but I haven't been getting one.Log in to reply
@Samarth Agarwal should post the next problem. – Harsh Shrivastava · 8 months ago
Okay according to rulesLog in to reply
– Samarth Agarwal · 8 months ago
Very nice..... Pls post the next questionLog in to reply
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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\( I_{n} = \displaystyle \int_{0}^{\infty} \dfrac{ \displaystyle \sum_{k=1}^{n} \left(\arctan\left(\dfrac{(k+1)x}{k}\right) \arctan(x) \right)}{x}dx \)
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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\(\displaystyle \int \dfrac{\cos x \ dx}{1+2 \sin x \cos x}\)
\(\displaystyle = \int \dfrac{\cos x \ dx}{(\sin x + \cos x )^2}\)
\(\displaystyle = \dfrac{1}{2} \int \dfrac{(\sin x + \cos x) + (\cos x  \sin x)}{(\sin x + \cos x )^2} dx\)
\( \displaystyle = \dfrac{1}{2} \int \dfrac{dx}{\sin x + \cos x} + \dfrac{1}{2} \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2}\)
\( \displaystyle = \dfrac{1}{2\sqrt{2}} \int \csc \left(x + \frac{\pi}{4}\right) dx + \dfrac{1}{2} \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2}\)
\( \displaystyle = \dfrac{1}{2\sqrt{2}} \ln \left \csc\left( x +\frac{\pi}{4}\right) + \cot \left(x+\frac{\pi}{4}\right) \right\dfrac{1}{2(\sin x +\cos x)} + C\) – Samuel Jones · 8 months ago
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– Adarsh Kumar · 8 months ago
My method was a little bit different!Although nice work!Log in to reply
– Aditya Sharma · 8 months ago
Yeah same form as @Adarsh Kumar , He just simplified the last one to \(\frac{1}{2\sqrt2}\frac{1}{sin(x+\frac{\pi}{4})}\)Log in to reply
@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
I thinkLog in to reply
– Adarsh Kumar · 8 months ago
Is the answer long?Log in to reply
– Aditya Sharma · 8 months ago
Not that really. Not as that of \(\sqrt{tanx}\)Log in to reply
– Adarsh Kumar · 8 months ago
Yeah not that long but does it contain log and some trigo functions?Log in to reply
– Aditya Sharma · 8 months ago
Yeah it has log & trigo functionsLog in to reply
– Adarsh Kumar · 8 months ago
Is it \[\dfrac{1}{2\sqrt{2}}\left(\ln \csc(x+\frac{\pi}{4})\cot(x+\frac{\pi}{4})\dfrac{1}{\sin(x+\frac{\pi}{4})}\right)+C\]Log in to reply
– Aditya Sharma · 8 months ago
Absolutely ! Great work.Log in to reply
– Adarsh Kumar · 8 months ago
Haha thanx!It was a good question!I am posting the solution!Log in to reply
– Aditya Sharma · 8 months ago
Yeah !Log in to reply
@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 should be allowed as he provided the solution a bit earlier. – Aditya Sharma · 8 months ago
I am a bit confused here, AccordinglyLog in to reply
@Akshay Yadav post the next note. I'll post Problem \(40\) there. – Samuel Jones · 8 months ago
Okay. LetLog in to reply
@Akshay Yadav please post the next note :) – Aditya Sharma · 8 months ago
Yeah thanx :P or else my phone would have exploded .Log in to reply
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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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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– Samarth Agarwal · 8 months ago
Absolutely correct!!! You may post the next question :)Log in to reply
PROBLEM 35:
Time for an easy problem!
Find the radius and coordinates of center of curvature of the curve
\[y=(4xx^23)^{\frac{1}{2}}\]
at the points \(x=1.2\) and \(x=2\). Also what is this curve? – Akshay Yadav · 8 months ago
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– Samarth Agarwal · 8 months ago
It is a semi circle above x axis as y is always positive with radius 1 unit and center (2,0)Log in to reply
– Akshay Yadav · 8 months ago
Wow, I had expected a fast reply but you did it extremely fast!Log in to reply
– Samarth Agarwal · 8 months ago
This is because it just requires simple co ordinate geometryLog in to reply
– Samarth Agarwal · 8 months ago
I think that question is more based on co ordinate geometry as curve is just a semi circleLog in to reply
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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\(\displaystyle \int_0^{\pi} \left(x^2+\pi^2a^2+\frac{b^2}{\pi^2}\cos^2 (x)2\pi ax+2ab\cos (x)\frac{2b}{\pi}\cos (x) \right) dx\)
From that you get the expression
\(\frac{3b^2+24b+6{\pi}^4a^26{\pi}^4a+2{\pi}^4}{6{\pi}}\)
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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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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@Samuel Jones Problem 40 on the new note please – Aditya Sharma · 8 months ago
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