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:
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Top NewestProblem 23 \[\large\int\dfrac{x^21}{x^3\sqrt{2x^42x^2+1}}\,dx=~?\]
This problem has been solved by Harsh Shrivastava.
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A similar problem like this came in JEE Mains 2016! I just saw the question paper and recalled this problem ;)
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Problem 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.
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Solution without using Beta function: \[\dfrac{5050\overbrace{\displaystyle \int_0^1(1x^{50})^{100}\,\mathrm{d}x}^{\color{red}{I_1}}}{\underbrace{\displaystyle \int_0^1(1x^{50})^{101}\,\mathrm{d}x}_{\color{red}{I_2}}}\]
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:
\[\dfrac{5050I_1}{I_2}=\boxed{5051}\]
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Both methods are same. The proof of Gamma functional equation uses IBP, so both methods are essentially the same.
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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 :
\(\large \int_{0}^{1} x^{m1}(1x)^{n1}=\beta(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\frac{(m1)!(n1)!}{(m+n1)!}\) 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 ,
\(\large \frac{5050.\int_{0}^{1}(1x^{50})^{100}dx}{\int_{0}^{1}(1x^{50})^{101}dx} = 5051\) \(\large \boxed{Proved}\)
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\(\large \mathbf{Problem}\) \(\large \mathfrak{27}\)
Prove that : \(\large \int_{0}^{\pi} \frac{xdx}{1+cos(\alpha)sin(x)} = \frac{\alpha\pi}{sin(\alpha)}\)
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\[\large I=\int_{0}^{\pi} \frac{xdx}{1+\cos(\alpha)\sin(x)}\]
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 \(\int \dfrac{dx}{x^2+a^2}=\frac 1a(\tan^{1} \frac xa)\) .
\[=\dfrac{\pi}{\sin \alpha}\left(\tan^{1}\left(\dfrac{t+\cos \alpha }{\sin \alpha}\right)\right)_0^{\infty}\]
\[=\dfrac{\pi}{\sin \alpha}(\dfrac{\pi}{2}\underbrace{\tan^{1} \left(\cot \alpha\right)}_{\dfrac{\pi}{2}\alpha})\]
\[\large \boxed{=\dfrac{\pi\alpha}{\sin \alpha}}\]
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Problem 24:
Evaluate \[\displaystyle \int \dfrac{dx}{(\ln^{3} (\tan x))(\sin 2x)} \]
This problem was solved by Rishabh Cool.
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Solution to Problem 24
Substitute \(\ln \tan x=t\) such that \(\dfrac{dx}{\sin 2x}=\dfrac{dt}{2}\). Integral transforms to: \[\int \dfrac{dt}{2 t^3}\] \[=\dfrac{1}{ 4t^2}=\dfrac{1}{4\ln^2 \tan x}+C\]
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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 :
\(f(4) = \frac{5(e^81)}{2e^4}\)
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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 \( f(4) = \displaystyle \sum_{n=1}^\infty \frac{ 4^n ( (1)^n + 5 ) }{ 8(n1)!} = \displaystyle \sum_{n=1}^\infty \frac{ 4^n (1)^n }{ 8(n1)!} + \frac{5}{2} \displaystyle \sum_{n=1}^\infty \frac{ 4^{(n1)} }{ (n1)!} = \frac{1}{8} \frac{4}{e^4} + \frac{5}{2} e^4 = \boxed{ \frac{ 5e^8  1}{2e^4} } \)
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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 \(f'(x) = \sqrt{5+{f(x)}^2}\)
\(\int dx = \int \frac{d({f(x)})}{\sqrt{5+{f(x)}^2}}\)
\(x + C_1 = logf(x) + \sqrt{5+{f(x)}^2}\)
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 ,
\(2f(x) = 5(e^xe^{x})\implies f(x) = \frac{5(e^xe^{x})}{2}\)
\(\large\boxed{ f(4) = \frac{5(e^4e^{4})}{2} = \frac{5(e^81)}{2e^4}}\)
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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.
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This is correct ^. In reality the function is exponential.
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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
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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 \(\dfrac{1}{p^2(p1)!}S\) 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.
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Problem 30
\[\large\displaystyle \int \dfrac{\mathrm{d}x}{x^{11}\sqrt{1+x^4}}=\ ?\]
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\( I = \displaystyle \int \frac{\mathrm{d}x}{x^{11}\sqrt{1+x^4}}= \)
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 , \( I =  \frac{ \sqrt{x^4 + 1}( 8x^8  4x^4 + 3)}{30x^{10}} + C \)
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Problem 29:
Prove that:
\[\displaystyle \lim_{x\to 0} \dfrac{24}{x^3} \displaystyle \int_0^x \dfrac{t \ln(1+t)}{t^4+4} dt =2 \]
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Applying LHopital:
\[\lim_{x\to 0}\dfrac{8x\ln(1+x)}{x^2(x^4+4)}\]
\[=8\lim_{x\to 0}\dfrac{\ln(1+x)}{x(4)}\]
\[\text{Using} \lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1\]
\[=\boxed 2\]
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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 \( \frac{I_{1}}{I_{2}}\) ?
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\( I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{2}} f(\sin 2x)\sin x dx \)
Using the property,
\( \displaystyle \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)+f(2ax)dx \)
\( I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x ) dx \)
In \( I_{2} \) ,
Use \( \displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b} f(a+bx)dx \)
\( \therefore I_{2} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)\cos\left( \dfrac{\pi}{4}  x\right) dx \)
\( \therefore I_{2} = \dfrac{1}{\sqrt{2}}\cdot \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x) dx = \dfrac{I_{1}}{\sqrt{2}} \)
\( \dfrac{I_{1}}{I_{2}} = \sqrt{2} \)
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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 \( I_{20}  I_{6} \)
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PROBLEM 33 \[\displaystyle \int \sqrt{\tan x} \ dx\]
The answer is very long so be patient.!
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\( \tan x = t^{2} \)
\( dx = \dfrac{2t}{1+t^{4}}dt \)
\( I = \displaystyle \int \dfrac{2t^{2}}{1+t^{4}}dt \)
\( I = \displaystyle \int \dfrac{t^{2}1}{1+t^{4}}dt + \int \dfrac{t^{2}+1}{1+t^{4}}dt \)
\( I = \displaystyle \int \dfrac{1\frac{1}{t^{2}}}{t^{2} + \dfrac{1}{t^{2}}}dt + \int \dfrac{1+\frac{1}{t^{2}}}{t^{2} + \dfrac{1}{t^{2}}}dt \)
\( I = \displaystyle \int \dfrac{1\frac{1}{t^{2}}}{\left(t+\frac{1}{t} \right)^{2}  2}dt + \int \dfrac{1+\frac{1}{t^{2}}}{\left(t\dfrac{1}{t}\right)^{2} + 2 }dt \)
\( I = \displaystyle \int \dfrac{du}{u^{2}  (\sqrt{2})^{2}} + \int \dfrac{dv}{v^{2}+(\sqrt{2})^{2}} \)
\( I =\dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{u\sqrt{2}}{u+\sqrt{2}}\right) + \dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{v}{\sqrt{2}}\right) \)
\( I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{t+\frac{1}{t}\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)+ \dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{t\frac{1}{t}}{\sqrt{2}}\right) \)
\( I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{\sqrt{\tan(x)} + \sqrt{\cot(x)}\sqrt{2}}{\sqrt{\tan(x)} + \sqrt{\cot(x)}+\sqrt{2}}\right)+\dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{\sqrt{\tan(x)}\sqrt{\cot(x)}}{\sqrt{2}}\right) + c \)
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Will you post next problem or shall we post? You often stall the contest.
(Sorry to be rude)
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@Samarth Agarwal should post the next problem.
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Very nice..... Pls post the next question
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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: \[f(2)=\dfrac{2e^2+1}{3}\]
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@Rishabh Cool is this your original problem??? Btw Its really very nice
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Nope... :).... I'm not sure but it might be an IIT JEE problem.... I found this problem interesting so I found it worth sharing.
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I don't have a good problem right now.... Can anybody else post the question... Thanx
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Sometimes I don't get differential equations, what should I do?
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Can i post the next question?
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@Akshay Yadav allows
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\(\text{Problem 38}\)\[\]What is the value of the integral given below?\[\int\dfrac{\tan x}{a+b\tan^2x}dx\]
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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
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\(\large I = \int \frac{tanx}{a+b(tan)^2x}\ = \int \frac{sin(x)cos(x)}{acos^2x+bsin^2x}\)
\(\text{Put } sin^2x=t\)
\(I = \frac{1}{2}\int \frac{sin(2x)}{a  sin^2x(ab)} = \int \frac{dt}{2(a  t(ab))} = \frac{1}{2(ba)}lnat(ab) + C\)
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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,
\( \displaystyle \int_{3}^{3e  \frac{1}{2}}g(x+1)dx \)
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\(I = \displaystyle \int_{3}^{3e  \frac{1}{2}} g(x+1) dx = \int_{4}^{3e+\frac{1}{2}} g(x)dx = \int_{f(0)}^{f(1)}g(x)dx \)
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,
\( I = \left[tf(t)\right]_{0}^{1}  \displaystyle \int_{0}^{1} f(t)dt \)
\( I = 3e + \dfrac{1}{2}  \displaystyle \int_{0}^{1} 3e^{t} + \dfrac{1}{1+t^{2}}dt = 3e + \dfrac{1}{2}  \left[ 3e^{t} + \tan^{1}(x) \right]_{0}^{1} \)
\( \therefore I = 3e + \dfrac{1}{2} 3e  \dfrac{\pi}{4} + 3 + 0 = \dfrac{7}{2}  \dfrac{\pi}{4} = \dfrac{14\pi}{4} \)
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PROBLEM 34:
Find the minimum value of the integral \(\displaystyle \int_0^{\pi} \left(x  \pi a  \frac {b}{\pi}\cos x\right)^2 \ dx \)
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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
\(\frac{3b^2+24b+6{\pi}^4a^26{\pi}^4a+2{\pi}^4}{6{\pi}}\)
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,
\[\boxed{\frac{\pi^3}{12}\frac{8}{\pi}}\]
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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?
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It is a semi circle above x axis as y is always positive with radius 1 unit and center (2,0)
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I think that question is more based on co ordinate geometry as curve is just a semi circle
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Wow, I had expected a fast reply but you did it extremely fast!
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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)
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\(\dfrac{f'(x)}{f(x)}=1\)
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???
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Absolutely correct!!! You may post the next question :)
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\(\mathbf{Problem} \mathfrak{39}\) Determine : \(\int \frac{dx}{secx+2sinx}\)
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The given integral can be written as
\(\displaystyle \int \dfrac{\cos x \ dx}{1+2 \sin x \cos x}\)
\(\displaystyle = \int \dfrac{\cos x \ dx}{(\sin x + \cos x )^2}\)
\(\displaystyle = \dfrac{1}{2} \int \dfrac{(\sin x + \cos x) + (\cos x  \sin x)}{(\sin x + \cos x )^2} dx\)
\( \displaystyle = \dfrac{1}{2} \int \dfrac{dx}{\sin x + \cos x} + \dfrac{1}{2} \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2}\)
\( \displaystyle = \dfrac{1}{2\sqrt{2}} \int \csc \left(x + \frac{\pi}{4}\right) dx + \dfrac{1}{2} \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2}\)
\( \displaystyle = \dfrac{1}{2\sqrt{2}} \ln \left \csc\left( x +\frac{\pi}{4}\right) + \cot \left(x+\frac{\pi}{4}\right) \right\dfrac{1}{2(\sin x +\cos x)} + C\)
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Yeah same form as @Adarsh Kumar , He just simplified the last one to \(\frac{1}{2\sqrt2}\frac{1}{sin(x+\frac{\pi}{4})}\)
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@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
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My method was a little bit different!Although nice work!
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Is the answer long?
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Not that really. Not as that of \(\sqrt{tanx}\)
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@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.
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@Samuel Jones should be allowed as he provided the solution a bit earlier.
I am a bit confused here, AccordinglyLog in to reply
@Akshay Yadav post the next note. I'll post Problem \(40\) there.
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@Akshay Yadav please post the next note :)
Yeah thanx :P or else my phone would have exploded .Log in to reply
Third note of this season has been posted. Click here to access it.
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@Samuel Jones Problem 40 on the new note please
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