$$ \int_0^{\pi/2} \frac{ \sqrt[3]{\tan x} }{ \left( \sin x + \cos x \right)^2 }\, dx $$
This question I use the Kings Property and I equal to cube root of $\tan x$ plus cube root of $\cot x$ divided by $(\sin x+\cos x)^2$ but was stuck after that.
Or if we pull out $\cos x$ from the denominator, then it will be $(\tan x+1)^2$ and then substituting $\tan x$ for $u$. I couldn’t proceed further...
Kindly help Thanks
Substitute $t={\tan x}$ \begin{align} &\int_0^{\pi/2} \frac{ \sqrt[3]{\tan x} }{ \left( \sin x + \cos x \right)^2 }\, dx\\ =&\int_0^\infty \frac {t^{1/3}}{(1+t)^2}dt\overset{ibp} = \int_0^\infty \frac {d(t^{1/3})}{1+t}\overset{t^{1/3}\to t}= \int_0^\infty \frac {dt}{1+t^3}=\frac{2\pi}{3\sqrt3} \end{align}
firstly we have $$\int_0^{\pi/2} \frac{\tan^{\frac{1}{3}}x}{(\sin x+\cos x)^2}dx=\int_0^{\pi/2} \frac{\tan^{\frac{1}{3}}x}{(\tan x+1)^2} \sec^2x dx $$ now put $x\to \tan^{-1} x$ and get $$\int_0^{\pi/2} \frac{\tan^{\frac{1}{3}}x}{(\sin x+\cos x)^2}dx=\int_0^{\infty} \frac{x^{\frac{1}{3}}}{(x+1)^2}dx $$ we can use the formula $$ \int_0^\infty \frac{x^n}{(x+1)^2}dx=\pi n \csc(\pi n) $$ So $$\int_0^{\pi/2} \frac{\tan^{\frac{1}{3}}x}{(\sin x+\cos x)^2}dx=\pi \frac{1}{3} \csc \left(\pi \frac{1}{3} \right)=\frac{2\pi}{3\sqrt{3}}$$ the formula we used can be easily prove by gamma function , but if you need to solve that integral without gamma function we can find indefinite integral firstly put $x\to \frac{1}{x}$ So $$ \int_0^{\infty} \frac{x^{\frac{1}{3}}}{(x+1)^2}dx=\int_0^{\infty} \frac{x^{\frac{-1}{3}}}{(x+1)^2}dx $$ So $$ \int_0^{\infty} \frac{x^{\frac{1}{3}}}{(x+1)^2}dx=\frac{1}{2} \int_0^{\infty} \frac{x^{\frac{1}{3}}+x^{\frac{-1}{3}}}{(x+1)^2}dx $$ now put $x \to x^3$ and get $$ \int_0^{\infty} \frac{x^{\frac{1}{3}}}{(x+1)^2}dx= \int_0^{\infty} \frac{3x^2(x+\frac{1}{x})}{(x^3+1)^2}dx $$ we can use partial fraction but its too long so lets see this trick $$\frac{3x^2(x+\frac{1}{x})}{(x^3+1)^2}=\frac{3x(x^2+1)}{(x^3+1)^2}-\frac{1+x}{x^3+1}+\frac{1+x}{x^3+1}$$ $$=\frac{-x^4+2x^3+2x-1}{(x^3+1)^2}+\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}$$ So $$\int_0^{\infty} \frac{x^{\frac{1}{3}}}{(x+1)^2}dx=\frac{1}{2}\int_0^\infty \frac{-x^4+2x^3+2x-1}{(x^3+1)^2} dx+\frac{1}{2}\int_0^\infty \frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}dx$$ first integral equals zero (using indefinite integral you can try with) and the second integral can be evaluate using arctan function $$ \frac{1}{2} \int_0^\infty \frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}dx=\left(\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2x-1}{\sqrt{3}} \right)\right)^\infty_0=\frac{2\pi}{3\sqrt{3}}$$ finally $$\int_0^{\pi/2} \frac{\tan^{\frac{1}{3}}x}{(\sin x+\cos x)^2}dx=\frac{2\pi}{3\sqrt{3}} $$
Substituting $$\tan x = t^3, \qquad \sec^2 x \,dx = 3 t^2 \,dt,$$ rationalizes the integral, giving $$3 \int_0^\infty \frac{t^3 \,dt}{(1 + t^3)^2},$$ and integrating by parts with $u = t$, $dv = \frac{t^2 \,dt}{(1 + t^3)^2}$, yields $$\int_0^\infty \frac{dt}{1 + t^3}.$$ To evaluate this latter integral, see Compute $\int_0^\infty \frac{dx}{1+x^3}$ or, if you're able to use residue calculus, Integrating $\int_0^{\infty} \frac{dx}{1+x^3}$ using residues..
