Brilliant Integration Contest - Season 2 (Part-1)

Hi Brilliant! Just like what Anastasiya Romanova conducted last year, this year I would also like to conduct an integration contest.

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:

• 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 integrals either definite or indefinite integrals.

• You are NOT allowed to post a multiple integrals problem.

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

• Do not copy questions from last year's contest. If anyone found to do so he/she will be banned from taking further part in this contest

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

The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, Dirichlet eta function, dilogarithm, digamma function, trigonometric integral, Wallis' integral, Bessel function, contour integration and Ramanujan's Master Theorem (including Mellin Transform).

Format your post is as follows:

 1 2 3 4 5 6 7 **SOLUTION OF PROBLEM xxx (number of problem) :** **[Post your solution here]** **PROBLEM xxx (number of problem) :** **[Post your problem here]** 

View Part 2

5 years, 6 months ago

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Problem 1:

Evaluate: $\int _{ 0 }^{ 1 }{\frac { x \ln\left( x \right) }{ \sqrt { 1-{ x }^{ 2 } } } \, dx }$

This problem has been solved by Surya Prakash.

- 5 years, 6 months ago

Another solution: Let $\sqrt{(1-x^2)}=t$

$\frac{-x}{\sqrt{(1-x^2)}}dx=dt$

So integration becomes

$- \int_1^0 ln\sqrt{(1-t^2)}dt$

$-\frac{1}{2} \int_1^0 ln(1-t^2)dt$

$\frac{1}{2} \int_0^1 ln(1-t^2)dt$

$\frac{1}{2} \int_0^1 (ln(1-t)+ln(1+t))dt$

This can be easily evaluated and comes out to be $ln2-1$

- 5 years, 6 months ago

$\Large \text{Solution to Problem 1:}$

$\int_{0}^{1} \dfrac{x \ln(x)}{\sqrt{1-x^2}} dx$ Take $t = \arcsin(x) \iff dt = \dfrac{dx}{\sqrt{1-x^2}}$ and the limits changes to $t=0$ to $t=\dfrac{\pi}{2}$. $\int_{0}^{\dfrac{\pi}{2}} \sin (t) \ln(\sin(t)) dt$

Consider,

$B(x,y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} dt$

Taking the transformation $t=\sin^2 (z) \iff dt = 2\sin(z) \cos (z)dz$, the integral transforms into

$B(x,y) = 2 \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) dz$

But $B(x,y) = \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)}$

So, $\int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) dz = \dfrac{1}{2} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)}$

Differentiate it w.r.t $x$

$2 \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) \ln(\sin(z)) dz = \dfrac{1}{2} \dfrac{\partial}{\partial x} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)} \\ \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) \ln(\sin(z)) dz = \dfrac{1}{4} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)} (\psi (x) - \psi(x+y))$

Take $x= 1$ and $y= \dfrac{1}{2}$.

$\int_{0}^{\pi / 2} \sin(z) \ln (\sin (z)) dz = \dfrac{1}{4} \dfrac{\Gamma (1) \Gamma(1/2) }{\Gamma (3/2)} (\psi (1) - \psi(3/2)) = \ln (2) -1$

- 5 years, 6 months ago

Problem 2:

Evaluate:

$\Large \int_{0} ^{\infty} \dfrac{x \cos (x^3)}{\exp (x^3)} \, dx$

This problem has been solved by Tanishq Varshney.

- 5 years, 6 months ago

$\large{Solution~to~Problem~ 2}$

Firstly I would add a proof of my solution

$\large{\displaystyle \int _{0}^{\infty} e^{-ax}x^{n-1} dx=\frac{\Gamma (n)}{a^{n}}}$ where $\Gamma (n)$ is Gamma function

Replace $a \rightarrow a+ib$ where $i =\sqrt{-1}$

$\large{\displaystyle \int _{0}^{\infty} e^{-ax}e^{-ibx}x^{n-1} dx=\frac{\Gamma (n)}{(a+ib)^{n}}}$

Put $a=r \cos y ~ and ~ b=r \sin y$.

So that $r^2=a^2+b^2$ and $y=\arctan \left(\frac{b}{a}\right)$

Using de moviers theorem

$\large{\displaystyle \int _{0}^{\infty} e^{-ax}(\cos bx-i \sin bx)x^{n-1} dx=\frac{\Gamma (n)}{r^{n}}(\cos ny+\sin ny)^{-1}}$

Comparing the real and imaginary parts we finally get

$\large{\displaystyle \int _{0}^{\infty}x^{n-1} e^{-ax}(\cos bx) dx=\frac{\Gamma (n)}{r^{n}}(\cos ny)}$

now in the given integral put $x^{3}=t$

The integral now becomes

$\large{\frac{1}{3} \displaystyle \int^{\infty}_{0} t^{-\frac{1}{3}}e^{-t} \cos (t) dt}$

here $n=\frac{2}{3}$;$a=1$ and $b=1$

Thus we get $\large{\frac{1}{3}\frac{\Gamma \left(\frac{2}{3}\right) \cos \left(\frac{\pi}{6}\right)}{2^{\frac{1}{3}}}}$

$\large{\boxed{\frac{\sqrt{3} \Gamma \left(\frac{2}{3}\right)}{3.2^{\frac{4}{3}}}}}$

- 5 years, 6 months ago

Problem 23:

Evaluate:$\int _{ 0 }^{ 1 }{ \log\left( 1+x \right) \log\left( 1-{ x }^{ 3 } \right) dx }$

Due to time constraint, Aditya Kumar decided to post a solution himself.

- 5 years, 6 months ago

Solution to Problem 23:

Substitute: $x\longrightarrow \frac { 1-x }{ 1+x }$

$I=\int_0^1 \ln(1+x)\ln(1-x^3)dx=2\int_0^1 \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right)\frac{dx}{(1+x)^2}$

On separating the integrands,

$\ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right) \\\small =\ln^22-\ln2\ln x+\ln2\ln(3+x^2)-4\ln2\ln(1+x)-\ln x\ln(1+x)+3\ln^2(1+x)-\ln(1+x)\ln(3+x^2)$

1st integral:

\int_0^1 \frac{dx}{(1+x)^2}=-\frac1{1+x}\Bigg{|}_0^1=\frac12

2nd integral:

$\int_0^1 \frac{\ln x}{(1+x)^2}dx=\sum_{n\geq1} \frac{(-1)^n}{n}=-\ln2$

3rd integral:

\int_0^1 \frac{\ln(1+x)}{(1+x)^2}dx=-\frac{\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac{dx}{(1+x)^2}dx=\frac12-\frac{\ln2}{2}

4th integral:

\int_0^1 \frac{\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2x}{(1+x)(3+x^2)}dx \\=-\ln2+\ln3+2\operatorname{Re} \int_0^1 \frac{dx}{(1+x)(x+i\sqrt{3})}=-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}

5th integral:

\int_0^1 \frac{\ln x \ln(1+x)}{(1+x)^2}dx=-\frac{\ln x\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac1{1+x}\left(\frac{\ln(1+x)}{x}+\frac{\ln x}{1+x}\right)dx \\=-\ln2+\int_0^1 \frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx=\frac{\pi^2}{12}-\frac12\ln^22-\ln2

6th integral:

\int_0^1 \frac{\ln^2(1+x)}{(1+x)^2}dx=-\frac{\ln^2(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2\ln(1+x)}{(1+x)^2}dx \\=1-\ln2-\frac12\ln^22

7th integral:

\int_0^1 \frac{\ln(1+x)\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(1+x)\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac1{1+x}\left(\frac{2x\ln(1+x)}{3+x^2}+\frac{\ln(3+x^2)}{1+x}\right)dx \\=-\ln^22-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}+2J

where,

$J=\int_0^1\frac{x\ln(1+x)}{(1+x)(3+x^2)}dx=\operatorname{Re} \int_0^1 \frac{\ln(1+x)}{(1+x)(x+i\sqrt{3})}dx \\=-\frac18\ln^22-\operatorname{Re} \,\frac1{i\sqrt{3}-1}\int_0^1 \frac{\ln(1+x)}{x+i\sqrt{3}}dx$

Now on using the formula:

$\int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2 \ln\frac{a+1}{a-1}+\operatorname{Li}_2\left(\frac2{1-a}\right)-\operatorname{Li}_2\left(\frac1{1-a}\right)$

Putting $a=i\sqrt{3}$ then gives, after taking the real part, and assuming the principle value of the logarithm,

$\small J=-\frac18\ln^22+\frac{\pi\sqrt{3}}{12}\ln2+\frac14\operatorname{Re}\text{Li}_2(e^{i\pi/3})-\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(e^{i\pi/3})-\frac14\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})+\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})$

Using:

$\displaystyle \,\, \operatorname{Re}\text{Li}_2(e^{i\pi/3})=\frac{\pi^2}{36}$,

$\displaystyle \,\, \operatorname{Im}\text{Li}_2(e^{i\pi/3})=\frac1{2\sqrt{3}}\psi_1\left(\frac13\right)-\frac{\pi^2}{3\sqrt{3}}$,

$\displaystyle \,\,\text{Li}_2(\frac12 e^{i\pi/3})=-\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\frac12\ln^2\left(\frac34-i\frac{\sqrt{3}}{4}\right)$,

and

$\displaystyle \,\, \operatorname{Re}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)^2\, 3^n}=\frac14\text{Li}_2\left(-\frac13\right)$

we have $\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi^2}{72}-\frac18\ln^2\left(\frac34\right)-\frac14\text{Li}_2\left(-\frac13\right)$

and $\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi}{12}\ln\left(\frac34\right)-\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).$

Finally, putting everything together gives: $I=\ln^22 - \frac18\ln^23 + 2\ln2\ln3 - \frac32\ln3 - 6\ln2 + 6-\frac{\pi}{4\sqrt{3}}(2 + \ln3) - \frac{37\pi^2}{72} + \frac12\psi_1\left(\frac13 \right ) - \frac14\text{Li}_2\left( -\frac13 \right ) + \sqrt{3}\Im\text{Li}_2\left( -\frac{i}{\sqrt{3}} \right )$

$...$

This was not original. I found it very awesome, so I thought to share it here.

- 5 years, 6 months ago

I don't know what's wrong with the rendering of the Latex, since it sometimes render and sometimes it doesn't. So I'm going to post in pictures the one's that didn't render. I think it's better to move this discussion to another note.

1st Integral:

3rd Integral:

4th Integral:

5th Integral:

6th Integral:

7th Integral:

- 5 years, 6 months ago

Whoops, I accidentally deleted Tanishq's comment:

Problem 5:

$\int_0^{\pi /2} \tan^{-1} (1729\cos^2 x) \, dx$

This problem has been solved by Sudeep Salgia.

- 5 years, 6 months ago

I'll post the outline of the solution as it is a bit cumbersome. Let
$\displaystyle I(a) = \int_0^{\frac{\pi}{2}} \arctan (a \cos^2 x) dx \Rightarrow I'(a) = \frac{\cos^2 x}{1 + a^2 \cos^4 x } dx = \frac{\sec^2 x}{\sec^4 x + a^2 } dx$ Substitute $\tan x = t$, to get $\displaystyle I'(a) = \int_0^{\infty} \frac{1}{t^4 + 2t^2 + (a^2 + 1)} dt$ . Put $a^2 + 1 = k^2$ and with some simple rearrangement, we get,
$\displaystyle 2k I'(a) = \int_0^{\infty} \frac{ 1+ \frac{k}{t^2}}{\left( t - \frac{k}{t} \right)^2 + 2(k+1)} - \int_0^{\infty} \frac{ 1- \frac{k}{t^2}}{\left( t + \frac{k}{t} \right)^2 + 2(-k+1)}$ Substitute $\displaystyle t - \frac{k}{t} = z$ and $\displaystyle t + \frac{k}{t} = y$ in the respective integrals and evaluate them easily (They come out to be $\arctan$ of something ). After all this we can write,
$\displaystyle I(a) = \frac{\pi}{2\sqrt{2}} \int \frac{1}{\sqrt{a^2 + 1}} \frac{1}{\sqrt{\sqrt{a^2 + 1} +1}} da$
This can be rearranged to be written as
$\displaystyle I(a) = \frac{\pi}{2\sqrt{2}} \int \frac{1}{1+ \frac{\sqrt{a^2 + 1} - 1 }{2}} \times \frac{1}{2\sqrt{ \frac{\sqrt{a^2 + 1} - 1 }{2} }} \times \frac{1}{2} \times \frac{a}{\sqrt{a^2 +1}} da$

This evaluates to $\displaystyle \frac{\pi}{2} \arctan \left( \sqrt{ \frac{\sqrt{a^2+1} - 1}{2}} \right)$

The constant of integration is $0$ from $I(0) = 0$. Put $a =1729$ to get the answer.

- 5 years, 6 months ago

Problem 24:

Prove that

$\large \int_0^{\pi /2} \dfrac1{\sin^8 x + \cos^8 x } \, dx = \dfrac{\sqrt{10-\sqrt2}}{2} \pi .$

This problem has been solved by Tanishq Varshney.

- 5 years, 6 months ago

$\text{Solution to Problem 24}$

$\large{\displaystyle \int ^{\infty}_{0} \frac{x^{m-1}}{1+x}dx=\frac{\pi}{\sin m \pi} \quad \quad 0

Refer to problem 18 for this step.

Now in the integral $\tan x \rightarrow t$

$\large{\displaystyle \int ^{\infty}_{0} \frac{(1+t^2)^3}{1+t^{8}}dt}$

$\large{t^8 \rightarrow z}$

$\large{\frac{1}{8} \left(\displaystyle \int ^{\infty}_{0} \frac{z^{\frac{7}{8}-1}}{1+z}+ \frac{3z^{\frac{5}{8}-1}}{1+z}+ \frac{3z^{\frac{3}{8}-1}}{1+z}+ \frac{z^{\frac{1}{8}-1}}{1+z} dz \right)}$

$\large{\rightarrow \frac{\pi}{4} \left(\frac{1}{\sin \left(\frac{\pi}{8} \right)}+\frac{3}{\cos \left(\frac{\pi}{8} \right)} \right)}$

$\large{=\frac{\sqrt{10-\sqrt{2}}}{2} \pi}$

- 5 years, 6 months ago

$\text{Problem 3}$

Evaluate:

$\large {\displaystyle \int^{\frac{\pi}{2}}_{0} \tan^{\frac{3}{5}} x \, dx}$

This problem has been solved by Aditya Kumar.

- 5 years, 6 months ago

Solution to Problem 3: $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( tanx \right) }^{ \frac { 3 }{ 5 } } } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( sinx \right) }^{ \frac { 3 }{ 5 } }{ \left( cosx \right) }^{ \frac { -3 }{ 5 } } } dx\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { B\left( \frac { 4 }{ 5 } ,\frac { 1 }{ 5 } \right) }{ 2 } ...........\left\{ B(x,y)\quad is\quad beta\quad function \right\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \Gamma \left( \frac { 4 }{ 5 } \right) \Gamma \left( \frac { 1 }{ 5 } \right) }{ 2\Gamma \left( 1 \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \pi }{ 2sin\left( \frac { \pi }{ 5 } \right) } \quad ...............\{ by\quad euler's\quad reflection\quad formula\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 2\pi }{ \sqrt { 10-2\sqrt { 5 } } }$

- 5 years, 6 months ago

Problem 4:

Let $P(x)$ be a polynomial such that $\frac { P\left( x+1 \right) }{ P\left( x \right) } =\frac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x+1 }$ and $P(2)=3$, evaluate$\int _{ 0 }^{ 1 }{ \left( { \tan }^{ -1 }\left( P(x) \right) .{ \tan }^{ -1 }\left( \sqrt { \frac { x }{ 1-x } } \right) \right) \, dx }.$

This problem has been solved by Tanishq Varshney.

- 5 years, 6 months ago

$\text{Solution to Problem 4}$

Clearly $P(x)=x^2-x+1$ and $P(1-x)=P(x)$

Also $\tan^{-1} (x)=\cot^{-1} \left(\frac{1}{x}\right)$

$\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}$

$\large{I=\displaystyle \int^{1}_{0} \left( \tan^{-1}(P(x)) \cot^{-1} \left( \sqrt{\frac{x}{1-x}}\right) \right) dx}$

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

$\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} - \displaystyle \int^{1}_{0} \tan^{-1}\left(\frac{1}{1+x(x-1)}\right)\right)}$

$\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} - \displaystyle \int^{1}_{0} \tan^{-1}\left(\frac{x-(x-1)}{1+x(x-1)}\right) dx \right)}$

$\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} -( \displaystyle \int^{1}_{0} \tan^{-1} (x) dx- \displaystyle \int^{1}_{0} \tan^{-1} (x-1) dx)\right)}$

$\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} -\left( \frac{\pi}{2}-2 \ln (2)\right)\right)}$

$\large{I=\frac{\pi}{2} \ln(2)}$

- 5 years, 6 months ago

Problem 9:

Evaluate the indefinite integral: $\large{\displaystyle \int \frac{x^2+20}{(x \sin x+5 \cos x)^2} \, dx}.$

This problem has been solved by Surya Prakesh.

- 5 years, 6 months ago

\begin{aligned} x^2 + 20 &= (x \sin(x) + 5 \cos(x))^2 + (5 \sin (x) - x \cos(x))^2 - 5 \\ &=(x \sin(x) + 5 \cos(x))^2 + (5 \sin (x) - x \cos(x))^2 - 5 \cos^2(x) - 5\sin^2 (x) -x \cos(x)\sin(x) + x\cos(x) \sin(x) \\ &= (x \sin(x) + 5 \cos(x)) ( 5 \cos(x) + x \sin(x) - \cos(x)) + (5 \sin(x) - x \cos(x))(5 \sin(x) - x \cos(x) - \sin(x)) \\ &= (x \sin(x) + 5 \cos(x)) ( 5 \cos(x) + x \sin(x) - \cos(x)) - (5 \sin(x) - x \cos(x))( x \cos(x) + \sin(x) - 5 \sin (x)) \\ &= (x \sin(x) + 5 \cos(x)) \dfrac{d(5 \sin(x) - x \cos(x))}{dx} - (5 \sin(x) - x \cos(x)) \dfrac{d(x \sin(x) + 5 \cos(x))}{dx} \end{aligned}

Let $u=x \sin(x) + 5 \cos(x)$ and $v=5 \sin(x) - x \cos(x)$. So, $x^2 + 20 = u\dfrac{dv}{dx} - v\dfrac{du}{dx} = u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}$.

So, $\int \dfrac{x^2 + 20}{(x \sin(x) + 5 \cos(x))^2} dx = \int \dfrac{ u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}}{u^2} dx= \dfrac{v}{u} = \dfrac{5 \sin(x) - x \cos(x)}{x \sin(x) + 5 \cos(x)}$

- 5 years, 6 months ago

Problem 13: (Originally posted by Tanishq)

Evaluate $\int_0^1 x \ln \left( \dfrac{1+x}{1-x} \right) \, dx.$

This problem has been solved by Surya Prakesh.

- 5 years, 6 months ago

By using integration by parts,

$\int x \ln \left(\dfrac{1+x}{1-x} \right) dx = \dfrac{x^2}{2} \ln \left(\dfrac{1+x}{1-x} \right) + x - \dfrac{1}{2} \ln \left(1+x \right) + \dfrac{1}{2} \ln \left(1-x \right)$

So,

\begin{aligned} \int_{0}^{1} x \ln \left(\dfrac{1+x}{1-x} \right) dx &= \lim_{a \rightarrow 0} \int_{0}^{1-a} x \ln \left(\dfrac{1+x}{1-x} \right) dx \\ &= \lim_{a \rightarrow 0} \dfrac{(1-a)^2}{2} \ln \left(\dfrac{2-a}{a} \right) + 1 - a - \dfrac{1}{2} \ln \left(2-a \right) + \dfrac{1}{2} \ln a \\ &= 1\end{aligned}

- 5 years, 6 months ago

Problem 6:
Show that for positive reals, $a , b$ where $a, $\int_a^b \arccos \left( \frac{x}{\sqrt{(a+b)x - ab}} \right) \, dx = \frac{\pi}{4} \frac{ (b-a)^2}{a+b}$

Statutory warning: The substitutions need not be very intuitive.

This problem has been solved Aditya Kumar.

- 5 years, 6 months ago

Solution to Problem 6:

$I=\int _{ a }^{ b }{ \text{arccos}\left( \frac { x }{ \sqrt { \left( a+b \right) x-ab } } \right) dx } \\ I=\int _{ a }^{ b }{ \text{arccos}\left( \frac { \frac { x }{ \sqrt { ab } } }{ \sqrt { \left( \frac { \left( a+b \right) }{ ab } \right) x-1 } } \right) dx } \\ \sqrt { \left( \frac { \left( a+b \right) }{ ab } \right) x-1 } =t\\ I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { 2t }{ a+b } arccos\left( \frac { \left( { t }^{ 2 }+1 \right) \sqrt { ab } }{ t } \right) \right) dt } \\ I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { 2t }{ a+b } arccos\left( \left( { t }+\frac { 1 }{ t } \right) \sqrt { ab } \right) \right) dt }$

Now using Integration By Parts,

$I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { { t }^{ 2 }+1 }{ { \left( \sqrt { \frac { b }{ a } } -\sqrt { \frac { a }{ b } } \right) }^{ 2 }-{ \left( { t }-\frac { 1 }{ t } \right) }^{ 2 } } \right) dt } \\ t=p+\sqrt { 1+{ p }^{ 2 } } \\ I=\int _{ \frac { 1 }{ 2 } \left( -\sqrt { \frac { b }{ a } } +\sqrt { \frac { a }{ b } } \right) }^{ \frac { 1 }{ 2 } \left( \sqrt { \frac { b }{ a } } -\sqrt { \frac { a }{ b } } \right) }{ \left( \frac { { 2p }^{ 2 } }{ \sqrt { \frac { \frac { a }{ b } +\frac { b }{ a } +2 }{ 4 } -{ p }^{ 2 } } } \right) dp }$

Now,

${ p }^{ 2 }=\left( \frac { \frac { a }{ b } +\frac { b }{ a } +2 }{ 4 } \right) sin\theta$

This can be evaluated easily.

$\therefore \quad I=\frac { \pi }{ 4 } \frac { { \left( b-a \right) }^{ 2 } }{ \left( a+b \right) }$

Moral of the question:
To whatever extent we learn anything, we must never forget our basics.
Note that this question is solved using very basic concepts.

Any elegant modification to the solution will be accepted.

- 5 years, 6 months ago

Problem 8:

Evaluate

$\large f(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a^2t^2}e^{i\omega t}dt$

This problem has been solved by both Tanishq Varshney (first) and Surya Prakash (second) almost at the same time.

- 5 years, 6 months ago

$\large{\text{Solution to problem 8}}$

$t \rightarrow -t$

adding this with the original equation

$\Large{2f(w)=\frac{2}{\sqrt{2 \pi}} \displaystyle \int^{\infty}_{-\infty}e^{-a^2 t^2} \cos wt dt}$

as $e^{iwt}+e^{-iwt}=2 \cos wt=2 \Re(e^{iwt})$

$\Large{f(w)=\frac{1}{\sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-a^2 t^2}e^{iwt} dt \right)}$

Let $at=z$

$\Large{f(w)=\frac{1}{a \sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-z^2}e^{i\frac{w}{a}z} dz \right)}$

$\Large{f(w)=\frac{1}{a \sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-(z-\frac{iw}{2a})^2}e^{-\frac{w^2}{4a^2}} dz \right)}$

Using Gaussian integral

$\Large{f(w)=\frac{1}{a \sqrt{2}} e^{-\frac{w^2}{4a^2}}}$

- 5 years, 6 months ago

$f(\omega) = \dfrac{1}{\sqrt{2 \pi}} \int_{- \infty} ^{\infty} e^{-a^2 t^2} e^{i\omega t}dt$

It implies that $f(\omega) = \big[\mathcal{F}(F) \big](\omega)$, where $F(t) = e^{-a^2 t^2}$.

Now, differentiate $F(t)$, we get $F'(t) = -2a^2 t F(t)$.

Applying fourier transformation on both sides gives $i \omega f(\omega) = -2i a^2 f'(\omega)$.

$\implies \omega f(\omega) = -2a^2 \dfrac{d f(\omega)}{d \omega} \\ \implies f(\omega) = c e^{-\dfrac{\omega^{2}}{4a^{2}}}$

Since $c = f(0) =\dfrac{1}{\sqrt{2 \pi}} \int_{- \infty} ^{\infty} e^{-a^2 t^2} dt = 2 \dfrac{1}{\sqrt{2 \pi}} \int_{0} ^{\infty} e^{-a^2 t^2} dt =\dfrac{1}{a \sqrt{2 \pi}} \Gamma(1/2) = \dfrac{1}{a\sqrt{2}}$.

Therefore, $f(\omega) =\dfrac{1}{a\sqrt{2}} e^{-\dfrac{\omega^{2}}{4a^{2}}}$.

- 5 years, 6 months ago

Problem 10:

Evaluate

$\large \int_{0}^{\pi/2} \dfrac{1}{(9 \tan^2 (x) + 16)^3} \, dx$

This problem has been solved by Pi Han Goh.

- 5 years, 6 months ago

Let $I$ denote the value of the integral. And let $3\tan (x) = 4\tan(y)$, differentiate with respect to $x$: $3\sec^2(x) = 4\sec^2(y) \left( \frac{dy}{dx} \right) \Rightarrow dx = \dfrac{4\sec^2(y)}{3\sec^2(x)} dy= \dfrac{12\sec^2(y)}{9\sec^2(x)} dy= \dfrac{12\sec^2(y)}{9\tan^2(x) + 9} dy= \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} dy$
And of course, $9\tan^2(x) + 16 = (3\tan(x))^2 + 16 = 16(\tan^2 (y) + 1) = 16\sec^2(y)$.
When $x = 0$, $y = 0$; when $x\to \frac\pi2^-$, $y \to \frac\pi2 ^-$ as well. The integral becomes
\begin{aligned} I &=& \int_0^{\pi/2} \dfrac{1}{(16\sec^2(y))^3} \cdot \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} \, dy \\ &=& \dfrac{12}{16^3} \int_0^{\pi/2} \dfrac1{\sec^4(y)} \cdot \dfrac1{16\tan^2 (y) + 9} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16\sin^2 (y) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16(1-\cos^2(y)) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16 - 7\cos^2(y)} \, dy. \\ \end{aligned}