# Brilliant Integration Contest - Season 1

I am interested in holding an Integration Contest here on Brilliant.org like any other online forums such as AoPS or Integrals and Series. The aims of the Integration Contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows

1. I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
2. 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.
3. Please make a substantial comment.
4. Make sure you know how to solve your own problem before posting it in case there is no one can answer it within a week, then you must post the solution and you have a right to post another problem.
5. 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.
6. The scope of questions is only computation of integrals either definite or indefinite integrals.
7. You are NOT allowed to post a multiple integrals problem as well as a complex integral problem.
8. You are also NOT allowed to post a solution using a contour integration or residue method.
9. The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, Dirichlet eta function, dilogarithm, digamma function, and trigonometric integral.

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :

PROBLEM xxx (number of problem) :

Please share this note so that lots of users here know this contest and take part in it. (>‿◠)✌

Okay, let the contest begin! Here is the first problem:

PROBLEM 1 :

For $a>0$, show that

$\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,dx=\pi\ln\left(\frac{a}{4}\right)$

P.S. You may also want to see Brilliant Integration Contest - Season 1 (Part 2) and Brilliant Integration Contest - Season 1 (Part 3).

Note by Anastasiya Romanova
6 years, 7 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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Here is the solution: $\displaystyle I = \int \limits_0^a \frac{ \text{ln } x }{\sqrt{ax - x^2}} \text{d}x$

$\displaystyle I = \int \limits_0^a \frac{ \text{ln } (a-x) }{\sqrt{ax - x^2}} \text{d}x$

$\displaystyle \Rightarrow 2I = \int \limits_0^a \frac{ \text{ln } (ax - x^2) }{\sqrt{ax - x^2}} \text{d}x$

Put $\displaystyle x = \frac{a}{2} (1 - \sin t ) \Rightarrow ax - x^2 = \frac{a^2}{4} \cos ^2 t$

Therefore, the integral becomes, $\displaystyle 2I = \int \limits_{\frac{\pi}{2}}^{\frac{-\pi}{2}} \frac{ \text{ln } (\frac{a^2}{4} \cos ^2 t ) }{\sqrt{\frac{a^2}{4} \cos ^2 t }} (\frac{a}{2} \cos t )(-\text{d}t)$

Rearranging, we obtain,
$\displaystyle 2I = 4\int \limits_0^{\frac{\pi}{2}} \text{ln } (\frac{a}{2} \cos t ) \text{d}t$

The value of $\displaystyle \int \limits_0^{\frac{\pi}{2}} \text{ln } ( \cos \theta ) \text{d}\theta$ is $\displaystyle \frac{ -\pi \text{ ln } 2 }{2}$ (which I have calculated separately and I can post if it is required).

Thus, we obtain, $\displaystyle I = \pi \text{ ln } \bigg( \frac{a}{2} \bigg) - 2\frac{ \pi \text{ ln } 2 }{2} = \pi \text{ ln } \bigg(\frac{a}{4} \bigg)$

- 6 years, 7 months ago

Find a closed form expression for the integral : $\displaystyle I = \int \sin (2015 x) \sin ^{2013} x \text{ d}x$

- 6 years, 7 months ago

$Solution \quad of \quad problem \quad 2$

Split $sin(2015x)$ as $sin(2014x+x)$ and our integral becomes :

$I=\displaystyle \int { ({ sin }^{ 2014 }x.cos(2014x)+sin(2014x)cos(x){ sin }^{ 2013 }(x))\quad dx }$

Multiply and divide it by $2014$ to get :

$\frac { 1 }{ 2014 } \displaystyle \int { ({ sin }^{ 2014 }x.(2014cos(2014x))+sin(2014x)(2014{ sin }^{ 2013 }(x)cos(x)))\quad dx }$

$\Rightarrow \displaystyle I=\frac { 1 }{ 2014 } \int { { sin }^{ 2014 }(x)dsin(2014x)+sin(2014x)(d{ sin }^{ 2014 }x) }$

$\Rightarrow \displaystyle I=\frac { 1 }{ 2014 } \int { d({ sin }^{ 2014 }(x)sin(2014x)) }$

$\large \displaystyle I=\frac { { sin }^{ 2014 }(x)sin(2014x) }{ 2014 } +C$

$\large Problem \quad 3$

Evaluate $I=\displaystyle \int _{ 0 }^{ \infty }{ { (\frac { sin(x) }{ x } })^{ 2 }dx }$

- 6 years, 7 months ago

Perfect. Nice work Ronak. Expecting a problem soon.

- 6 years, 7 months ago

Nicely done, +1. Now you must post a new problem, but before that. I think it would be nice if you post this as a new post, no need to reply @Sudeep Salgia. See the rules & the format post. Thank you. :)

- 6 years, 7 months ago

Where is your proposed problem? We are waiting it. Please don't let us wait too long like this. You should post your solution and your proposed problem at once.

- 6 years, 7 months ago

@Anastasiya Romanova I have posted my problem sorry for holding the contest.

- 6 years, 7 months ago

It's okay. I have posted the solution of your problem and also proposed a new problem. :)

- 6 years, 7 months ago

Nicely done, +1! But you forget to post your problem anyway. You have a privilege for that so that this contest keeps going on. Post PROBLEM 2 in your thread solution. Thanks :)

- 6 years, 7 months ago

Well, there must be a rule for people like Ronak! Solving a problem doesn't give you the right to withhold the contest. In such a case, someone should be allowed to interfere, right?

- 6 years, 7 months ago

Sorry If that troubled all because actually I have to go to coaching centre, and my internet connection got down and I went to my coaching centre, I have returned just now and has posted the question.

- 6 years, 7 months ago

OK, if no-one posts his/her own problem after solving a problem within a day. The one who has a right to post a problem is the last solver before him/her. Is this fair?

- 6 years, 7 months ago

Solution of Problem 3 :

Use integration by parts by taking $u=\sin^2x$ and $dv=\dfrac{dx}{x^2}$, we get

\begin{aligned} \int_0^\infty\frac{\sin^2x}{x^2}\,dx&=-\left.\frac{\sin^2x}{x}\right|_0^\infty+\int_0^\infty\frac{2\sin x\cos x}{x}\,dx\\ &=0+\int_0^\infty\frac{\sin 2x}{x}\,dx\\ &=\int_0^\infty\frac{\sin t}{t}\,dt\qquad\Rightarrow\qquad t=2x\\ \end{aligned}

Now consider

$I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt\qquad a\ge0$

so that $I(\infty)=0$ and our considered integral is $I(0)$. Differentiating w.r.t. $a$ and then integrating back, we get

\begin{aligned} I'(a)&=-\int_0^\infty e^{-at}\sin t\,dt\\ &=-\frac{1}{1+a^2}\qquad\Rightarrow\qquad\text{integration by parts twice}\\ I(a)&=-\int \frac{1}{1+a^2}\,da\\ &=-\arctan(a)+C \end{aligned}

For $I(\infty)=0$, implying $C=\dfrac{\pi}{2}$. Hence

$I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt=\frac{\pi}{2}-\arctan(a)$

and

$I(0)=\int_0^\infty\frac{\sin t}{t}\,dt=\frac{\pi}{2}$

Thus

$\int_0^\infty\frac{\sin^2x}{x^2}\,dx=\frac{\pi}{2}$

Problem 4 :

Prove

$\int_0^{\pi/2}\frac{\ln(\cos x)}{\sin x}\,dx=-\frac{\pi^2}{8}$

- 6 years, 7 months ago

$\large Solution\quad of\quad Problem\quad 4$

Firstly we will prove a general result

Result

$\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a }ln(x)dx } =\frac { -1 }{ { (1+a) }^{ 2 } }$

Proof

$I=\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a }ln(x)dx }$

Integrating by parts $u=ln(x),dv={x}^{a}dx$

$I=\displaystyle ln(x)\frac { { x }^{ a+1 } }{ a+1 } \overset { 1 }{ \underset { 0 }{ | } } -\frac { 1 }{ a+1 } \int _{ 0 }^{ 1 }{ { x }^{ a }dx }$

$I=\frac { -1 }{ { (1+a) }^{ 2 } }$

Hence proved

Now we have :

$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { ln(cos(x)) }{ sin(x) } dx }$

Take $cos(x)=y$ to get our integral as :

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(y)dy }{ 1-{ y }^{ 2 } } }$

Now $\frac { 1 }{ 1-{ y }^{ 2 } } =\displaystyle \sum _{ n=0 }^{ \infty }{ { y }^{ 2n } }$

Our integral becomes :

$I=\displaystyle \sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ { y }^{ 2n }ln(y)dy } }$

Using our proved result we get :

$I=-\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { (2n+1) }^{ 2 } } }$

Also the summation can we written as :

$I=-(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } } -\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2n) }^{ 2 } } } )$

$I=-(\zeta (2)-\frac { 1 }{ 4 } \zeta (2))=\frac { -3 }{ 4 } \zeta (2)=\large \frac{-{\pi}^{2}}{8}$

Where $\zeta(x)$ is the zeta function

$\large Problem \quad 5$

Find closed form of $\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ n } } }$

- 6 years, 7 months ago

Keep going fella, +1! But please post your solution & proposed problem in a new thread like I did. No need to reply other threads to post a solution & a problem. Anyway, I've posted the solution of your problem & also proposed a new problem.

- 6 years, 7 months ago

Solution of Problem 7:

Proposition :

\begin{aligned} \int_0^{\pi} \frac{dx}{p+\cos x}\,dx=\frac{\pi}{\sqrt{p^2-1}} \end{aligned}

Proof :

It can be proven by using Weierstrass substitution: $t=\tan\left(\dfrac{x}{2}\right)$, then

\begin{aligned} \\ \int_0^{\pi} \frac{dx}{p+\cos x}\,dx &=\int_0^{\infty} \frac{2}{p+1+(p-1)t^2}\,dt\qquad\Rightarrow\qquad t=\sqrt{\frac{p+1}{p-1}}\tan y\\ &=\left.\frac{2}{\sqrt{p^2-1}}\arctan t\;\right|_0^{\infty}\\ &=\frac{\pi}{\sqrt{p^2-1}}\qquad\qquad\text{Q.E.D.} \end{aligned}

Differentiating the proposition w.r.t. $p$ twice and setting $p=\sqrt{10}$, we have

\begin{aligned} \\ \frac{\partial^2}{\partial p^2}\int_0^\pi\frac{1}{p+\cos x}\, dx&=\frac{\partial^2}{\partial p^2}\left[\frac{\pi}{\sqrt{p^2-1}}\right]\\ \int_0^\pi\frac{2}{\left(p+\cos x\right)^3}\, dx&=\frac{\pi\left(2p^2+1\right)}{\sqrt{\left(p^2-1\right)^5}}\\ \int_0^\pi\frac{1}{\left(\sqrt{10}+\cos x\right)^3}\, dx&=\frac{7\pi}{162} \end{aligned}

Problem 8 :

Prove

$\int_0^{\Large\frac{\pi}{2}}\frac{dx}{1+8\sin^2(\tan x)}=\frac{\pi}{6}\left(\frac{2e^2+1}{2e^2-1}\right)$

- 6 years, 7 months ago

Solution of Problem 8:

Substitute $\tan x\mapsto x$, then the integral is:

$\int_0^{\infty} \frac{dx}{(1+x^2)(1+8\sin^2x)}=\int_0^{\infty} \frac{dx}{(1+x^2)(5-4\cos(2x))}$

$=\int_0^{\infty} \frac{1}{1+x^2}\left(\frac{1}{3}\left(1+2\sum_{k=1}^{\infty} \frac{\cos(2kx)}{2^k}\right)\right)=\frac{\pi}{6}+\frac{2}{3}\sum_{k=1}^{\infty} \frac{1}{2^k}\int_0^{\infty} \frac{\cos(2kx)}{1+x^2}\,dx$

$=\frac{\pi}{6}+\frac{\pi}{3}\sum_{k=1}^{\infty} \left(\frac{1}{2e^2}\right)^k=\frac{\pi}{6}+\frac{\pi}{3}\frac{1}{2e^2-1}=\boxed{\dfrac{\pi}{6}\left(\dfrac{2e^2+1}{2e^2-1}\right)}$

I have used the following result:

$\int_0^{\infty} \frac{\cos(mx)}{x^2+a^2}\,dx=\frac{\pi}{2a}e^{-am}$

I cannot think of a challenging problem at the moment, I request somebody else to post one. Thanks!

- 6 years, 6 months ago

Ingenious! But I have a doubt - In your fourth step, how is it guaranteed that the function $\dfrac{\cos (2kx)}{2^k}$ is positive on $(0,\infty)$? PS : you forgot to write the $\mathrm{d}x$ in the third step.

- 6 years, 6 months ago

I don't know what did you mean by positive/negative on $(0,\infty)$, but if you meant to swap between integral & summation sign, it is valid because $\frac{\cos(2kx)}{1+x^2}$ is continuous, finite & integrable, therefore swapping those two signs can be justified by Fubini's theorem.

- 6 years, 6 months ago

If pranav refuses to put up a question then who put will put it up @Anastasiya Romanova .

- 6 years, 6 months ago

According to the rule 4, the one is the last solver and it turns out that one is me. See PROBLEM 9 below, I've just proposed it.

- 6 years, 6 months ago

Right. I confused it with Tonelli's theorem.

- 6 years, 6 months ago

Sorry for very late response. Using elementary techniques,-

Solution - Problem 7

$\displaystyle \int_{0}^{\pi} \dfrac{dx}{p - cosx}$

$2I = \displaystyle \int_{0}^{\pi} \dfrac{2p~dx}{p^2 - cos^2x}$

$I = \displaystyle \int_{0}^{\pi} \dfrac{p~dx}{p^2(1 + tan^2x) - 1}$

$I = p \displaystyle \int_{0}^{\pi} \dfrac{sec^2x~dx}{p^2tan^2x + \sqrt{(p^2 - 1)^2}}$

tanx = t

$I = 2p \displaystyle \int_{0}^{\infty} \dfrac{dt}{p^2t^2 + \sqrt{(p^2 - 1)^2}}$

$I = \dfrac{2p}{ \sqrt{p^2 - 1}} \times \dfrac{1}{p} \Big[_{0}^{\infty} tan^{-1} \dfrac{pt}{ \sqrt{p^2 - 1}}$

$I = \dfrac{\pi}{ \sqrt{p^2 - 1}}$

- 6 years, 5 months ago

Solution to problem 10

First integrate by parts -

$I= \int_{0}^{\infty} \dfrac{\sin^3{x}}{x^2} dx =\int_{0}^{\infty} \dfrac{3\sin^2{x} \cos{x}}{x} dx$

Then, use the property of the laplace transform that $\mathcal{L} \left\{\dfrac{f(t)}{t}\right\}(s)=\displaystyle\int_{s}^{\infty} F(p) dp$ (in our case, $s \rightarrow 0$).

$I=3\int_{0}^{\infty} \mathcal{L} \{\sin^2{t} \cos{t}\}(p) \ dp$

$I=3\int_{0}^{\infty} \dfrac{2s}{(s^2+1)(s^2+9)} ds=\boxed{\dfrac{3\log {3}}{4}}$ The last integral can be found by using partial fractions.

Note : $\mathcal{L}\{f(t)\}(s)=\displaystyle\int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t$

Problem 11

Find

$\int_{0}^{\infty} \dfrac{\log {x}}{x^2+\alpha^2} \mathrm{d}x \ \ \ \ \ \text{for} \ \alpha>0$

- 6 years, 6 months ago

Solution of Problem 13 :

First we prove the following proposition:

Proposition :

$\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}} \qquad\text{for }\ k=0,1,2,\ldots$

Proof :

Note that $\int_0^1 x^a\ dx=\frac1{a+1}\qquad\text{for }\ \alpha>-1.$ Differentiating equation above $k$ times w.r.t. $a$ we have $\int_0^1 \frac{\partial^k}{\partial a^k}\left(x^a\right)\ dx=\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}}\qquad\text{Q.E.D.}$

Now, we will evaluate the general case of

$\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\int_0^\infty\frac{x^{a-1}\,e^{-bx}}{1-e^{-bx}}\,dx\qquad\text{for }\ a,b>0$

Set $y=e^{-bx}$, then

$\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\frac{(-1)^{a-1}}{b^a}\int_0^1\frac{\ln^{a-1}y}{1-y}\,dy$

Use a geometric series for $\dfrac{1}{1-y}$ then interchange the integral and summation sign which is justified by the Fubini–Tonelli theorem and apply the previous proposition, we have

\begin{aligned} \int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx&=\frac{(-1)^{a-1}}{b^a}\int_0^1\sum_{k=0}^\infty y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty\int_0^1 y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty \frac{(-1)^{a-1}\, (a-1)!}{(k+1)^{a}}\\ &=\frac{(a-1)!}{b^a}\sum_{k=1}^\infty \frac{1}{k^{a}}\\ &=\frac{\Gamma(a)\,\zeta(a)}{b^a} \end{aligned}

where $\Gamma(a)$ is the gamma function and $\zeta(a)$ is the Riemann zeta function.

Hence, by setting $a=n+1$ and $b=1$, we have

$\int_0^\infty\frac{x^{n}}{e^{x}-1}\,dx=\Gamma(n+1)\,\zeta(n+1)$

Problem 14 :

Prove that

\begin{aligned} \int_{0}^1 \ln\left(\frac{3+x}{3-x}\right)\,\frac{dx}{\sqrt{x(1-x)}}=\pi\ln\left(\dfrac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\\ \end{aligned}

- 6 years, 6 months ago

Since @Pranav Arora is unable to propose a problem (I hope it's only temporary) and to make this contest sustains, then according to rule 4, I, as the last solver, have a right to propose a new one. Here is the problem:

PROBLEM 9

Prove that

$\int_0^1 \ln\left(\frac{1+x}{1-x}\right)\frac{dx}{x\sqrt{1-x^2}}=\frac{\pi^2}{2}$

- 6 years, 6 months ago

Solution -Problem - 9 We substitute, $t = \frac{1+x}{1-x} \implies x = \frac{1-t}{1+t}$ Doing the substitution and , simplifying , gives, $I = - \int_{0}^{1} \frac{\ln(t)}{(\sqrt{t})(1-t)} dt$ Now we substitute, $t=\sin^2(x)$ $I = -4\int_{0}^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cos(x)} dx$ Using, Problem -$4$ $I = -4 \times \frac{-\pi^2}{8} = \frac{\pi^2}{2}$

- 6 years, 6 months ago

I had already posted the solution to the problem and a new problem also you should not reply instead post your solution as a seperate comment as I did @Shivang Jindal

- 6 years, 6 months ago

Lol, time difference of 5mins. I think i was typing solution when you posted the solution .

- 6 years, 6 months ago

You can try my posted question. @Shivang Jindal

- 6 years, 6 months ago

I am able to write the integral as a series - $I=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}$

- 6 years, 6 months ago

$Solution\quad of\quad problem\quad 9$

$\displaystyle I=\int _{ 0 }^{ 1 }{ ln(\frac { 1+x }{ 1-x } )\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }$

Put $x=cos(\theta)$ to get our integral as :

$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln({ cot }^{ 2 }\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } }$

$\Rightarrow \displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(tan\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } }$

Put $tan(\theta/2)=x$ to get our integral as :

$\displaystyle (-4)\int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } }$

I proved in my solution to problem 4 that :

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } =\frac { -{ \pi }^{ 2 } }{ 8 }$

Using this I get :

$I=\frac{{\pi}^{2}}{2}$

$Problem\quad 10$

Find $\displaystyle \int _{ 0 }^{ \infty }{ \frac { { sin }^{ 3 }x }{ { x }^{ 2 } } dx }$

- 6 years, 6 months ago

Solution of Problem 11 :

Using substitution $u=\dfrac{\alpha^2}{x}\;\Rightarrow\;x=\dfrac{\alpha^2}{u}\;\Rightarrow\;dx=-\dfrac{\alpha^2}{u^2}\ du$, then

\begin{aligned} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\int_0^\infty\frac{\ln \left(\dfrac{\alpha^2}{u}\right)}{\left(\dfrac{\alpha^2}{u}\right)^2+\alpha^2}\cdot \dfrac{\alpha^2}{u^2}\ du\\ I(\alpha)&=\int_0^\infty\frac{2\ln \alpha-\ln u}{\alpha^2+u^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-\int_0^\infty\frac{\ln u}{u^2+\alpha^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-I(\alpha)\\ I(\alpha)&=\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du. \end{aligned}

The last integral can easily be evaluated by using substitution $u=\alpha\tan\theta$, then

\begin{aligned} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\frac{\ln \alpha}{\alpha}\int_0^{\Large\frac\pi2} \ d\theta\\ &=\frac{\pi\ln \alpha}{2\alpha} \end{aligned}

Problem 12 :

Prove

$\int_{-\infty}^{\infty}\frac{\cos \left(\, \arctan 2x\right)}{(1+x^2)\sqrt{1+4x^2}}\,dx=\frac{\pi}{3}$

- 6 years, 6 months ago

The last substitution should be $u=\alpha \tan \theta$ :)

- 6 years, 6 months ago

Fixed it. Sorry, I'm too hasty. Thanks... :)

- 6 years, 6 months ago

This is a simple kind of NCERT problem. It can be easily done by substituting $t = tan^{-1}\dfrac{x}{\alpha}$ and then applying by parts

- 6 years, 5 months ago

$Solution\quad of\quad Problem\quad 12$

It's a very good disguised integral.

First note that the function is an even function hence our integral can be written as :

$\displaystyle I=2\int _{ 0 }^{ \infty }{ \frac { cos(arctan(2x)) }{ (1+{ x }^{ 2 })\sqrt { 1+4{ x }^{ 2 } } } dx }$

Now we know that $\displaystyle cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}$. Using this our integral becomes :

$\displaystyle I=2\int _{ 0 }^{ \infty }{ \frac { 1 }{ (1+{ x }^{ 2 })(1+4{ x }^{ 2 }) } dx }$

Splliting it by partial fractions we get :

$\displaystyle I=\frac { 2 }{ 3 } (\int _{ 0 }^{ \infty }{ \frac { dx }{ (\frac { 1 }{ 4 } +{ x }^{ 2 }) } } -\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ 2 } } } )$

$\displaystyle I=\frac { 2 }{ 3 } (2{ tan }^{ -1 }(2x)\overset { \infty }{ \underset { 0 }{ | } } -{ tan }^{ -1 }(x)\overset { \infty }{ \underset { 0 }{ | } } )$

$\displaystyle I=\frac { \pi }{ 3 }$

$Problem\quad 13$

Find closed form of $\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx }$

- 6 years, 6 months ago

Wait a sec!? Could you rectify your solution, because the original problem is written as $\arctan(2x)$. And one thing, what did you mean by "Now we know that $\cos(\arctan x)=\frac{1}{\sqrt{1+4x^2}}$"? Could you elaborate?

- 6 years, 6 months ago

That was my typing mistake, also Let$arctan(2x)=\theta$, hence $2x=tan(\theta)$, hence $cos(\theta)=\frac{1}{\sqrt{1+4{x}^{2}}}$

$\Rightarrow cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}$

- 6 years, 6 months ago

$Solution\quad of\quad problem\quad 14$

Lemma

$\displaystyle\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx } =2\pi ln(\frac { a+b }{ 2 } )$

Proof :

$\displaystyle I(a)=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx }$

Differentiating with respect to $a$ we get :

$\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 2 }x }{ { a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x } dx }$

Put $tan(x)=t$ to get our integral as :

$\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } }$

$\displaystyle =2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } } =\frac { 2a }{ { b }^{ 2 }-{ a }^{ 2 } } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { b }^{ 2 }dt }{ { a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 } } } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ 1+{ t }^{ 2 } } } )$

Solving it we get :

$\displaystyle {I}^{'}(a)=\frac { 2\pi }{ a+b }$

$\Rightarrow \displaystyle I(a)=2\pi ln(a+b)+C$

Put $a=b=1$ to get $C=-2\pi ln(2)$

Hence $\displaystyle I(a)=2\pi ln(\frac{a+b}{2})$

In our integral in the given question put $x=\frac{1-sin\theta }{2}$ to get the integral as :

$I= \displaystyle \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7-sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5+sin(\theta))dx }$

Using the property $\displaystyle \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(a+b-x)dx }$ we get :

$\displaystyle I=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7+sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5-sin(\theta))dx }$

Adding these two we get :

$\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49-{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25-{ sin }^{ 2 }(\theta))dx } )$

$\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49{ cos }^{ 2 }\theta+48{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25{ cos }^{ 2 }\theta+24{ sin }^{ 2 }\theta)dx } )$

Using our given lemma we get :

$\displaystyle I=\frac { 1 }{ 2 } (2\pi ln(\frac{7+4\sqrt { 3 }}{2} )-2\pi ln(\frac{5+2\sqrt { 6 }}{2} ))=\pi ln(\frac { 7+4\sqrt { 3 } }{ 5+2\sqrt { 6 } } )$

$Problem\quad 15$

Evaluate $\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ ln(x) } +\frac { 1 }{ 1-x } dx }$

- 6 years, 6 months ago

According to timeline, you're faster than @jatin yadav to post your solution, then the right to propose a new problem is yours.

- 6 years, 6 months ago

_ Solution of problem 14 _

$I = \displaystyle \int_{0}^{1} \ln \bigg(\dfrac{3+x}{3-x} \bigg) \dfrac{1}{\sqrt{x(1-x^2)}} {\mathrm dx}$

= $\displaystyle \int_{0}^{1} \ln(1 + x/3) \dfrac{1}{\sqrt{x(1-x)}} {\mathrm dx}$

= $\displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 + \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta - \displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 - \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta$

Lemma $\displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 \theta) = \pi \ln \bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)$

Proof

Let $I(a) = \displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 x) {\mathrm dx}$