Brilliant Integration Contest - Season 2 (Part-1)

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 $ln2-1$

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

$\int_{0}^{1} \dfrac{x \ln(x)}{\sqrt{1-x^2}} dx$ 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 $y= \dfrac{1}{2}$.

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

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

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

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

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.

$\text{Solution to Problem 4}$

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

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

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

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

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

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

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

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

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

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

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.

$\text{Solution to Problem 24}$

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

Refer to problem 18 for this step.

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

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

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

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

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

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

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

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

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

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,

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

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.

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

$t \rightarrow -t$

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

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

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

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

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

$\begin{aligned} I &=& \int_0^{\pi/2} \dfrac{1}{(16\sec^2(y))^3} \cdot \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} \, dy \\ &=& \dfrac{12}{16^3} \int_0^{\pi/2} \dfrac1{\sec^4(y)} \cdot \dfrac1{16\tan^2 (y) + 9} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16\sin^2 (y) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16(1-\cos^2(y)) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16 - 7\cos^2(y)} \, dy. \\ \end{aligned}$