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

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

Solution:

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}

Team Gmath

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