evaluate int 0 pi 2 frac xsin x cos x sin4x cos4x

Question 1:

Evaluate: $\displaystyle\int_0^{\pi/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$

Let I = $\displaystyle\int_0^{\pi/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$ . . . (i)

Using property

$\int_0^a f(x) dx = \int_0^af(a - x) dx$

I = $\displaystyle\int_0^{\pi/2} \frac{(\pi/2-x)\sin (\pi/2 -x)\cos (\pi/2-x)}{\sin^4 (\pi/2 - x)+\cos^4 (\pi/2-x)}dx$

⇒ I = $\displaystyle\int_0^{\pi/2} \frac{(\pi/2-x)\cos x\sin x}{\cos^4 x+\sin^4 x}dx$

⇒ I = $\displaystyle \dfrac{\pi}{2}\int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4 x+\cos^4 x}dx-\int_0^{\pi/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$

⇒ $I = \displaystyle \dfrac{\pi}{2}\int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4 x+\cos^4 x}dx - I$

$\Rightarrow 2I = \displaystyle \dfrac{\pi}{2}\int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4 x+\cos^4 x}dx$

Dividing by $\cos^4 x$ in numerator and denominator

⇒ $2I = \displaystyle \dfrac{\pi}{2}\int_0^{\pi/2} \frac{\frac{\sin x\cos x}{\cos^4 x}}{\frac{\sin^4 x+\cos^4 x}{\cos^4 x}}dx$

$\Rightarrow 2I = \displaystyle \dfrac{\pi}{2}\int_0^{\pi/2} \frac{\tan x\sec x}{\tan^4 x+1}dx$

$\Rightarrow 2I = \displaystyle \dfrac{\pi}{2}\int_0^{\pi/2} \frac{\tan x\sec x}{(\tan^2 x)^2+1}dx$ ……..(ii)

Let $\tan^2 x = t$

$\Rightarrow 2\tan x\sec^2 x = \frac{dt}{dx}$

$\Rightarrow dx = \frac{dt}{2\tan x\sec^2 x}$

The limits are, when $x = 0, t = 0; x=\frac{\pi}{2}, t = \infty$

$2I = \displaystyle \dfrac{\pi}{2}\int_0^{\infty} \frac{\tan x\sec x}{(t)^2+1}\frac{dt}{2\tan x\sec^2 x}$

$\Rightarrow 2I = \displaystyle \dfrac{\pi}{4}\int_0^{\infty} \frac{dt}{(t)^2+1}$

$\Rightarroiw 2I = \displaystyle\dfrac{\pi}{4}[\tan^{-1}t]_0^{\infty}$

$\Rightarrow 2I = \dfrac{\pi}{4}[\tan^{-1}\infty - \tan^{-1}0]$

$\Rightarrow 2I = \dfrac{\pi}{4}[\dfrac{\pi}{2}-0]$

$\Rightarrow 2I = \dfrac{\pi^2}{8}$

$\Rightarrow I = \dfrac{\pi^2}{16}$