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Find the integration from 0 to pi/2 of [xsinx cosx/sin 4 x + cos 4 x]

Find the integration from 0 to pi/2 of
[xsinx cosx/sin4x + cos4x]

Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:Hello student, please find answer to your question
I = \int_{0}^{\pi /2}\frac{xsin(x)cos(x)}{sin^{4}(x) + cos^{4}(x)}dx
\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx
I = \int_{0}^{\pi /2}\frac{(\frac{\pi }{2}-x)sin(\frac{\pi }{2}-x)cos(\frac{\pi }{2}-x)}{sin^{4}(\frac{\pi }{2}-x) + cos^{4}(\frac{\pi }{2}-x)}dx
I = \int_{0}^{\pi /2}\frac{(\frac{\pi }{2}-x)cos(x)sin(x)}{cos^{4}(x) + sin^{4}(x)}dx
I = \int_{0}^{\pi /2}\frac{(\frac{\pi }{2})cos(x)sin(x)}{cos^{4}(x) + sin^{4}(x)}dx-\int_{0}^{\pi /2}\frac{(x)cos(x)sin(x)}{cos^{4}(x) + sin^{4}(x)}dx
I = \int_{0}^{\pi /2}\frac{(\frac{\pi }{2})cos(x)sin(x)}{cos^{4}(x) + sin^{4}(x)}dx-I
2I = \frac{\pi }{2}\int_{0}^{\pi /2}\frac{(1)cos(x)sin(x)}{cos^{4}(x) + sin^{4}(x)}dx
I = \frac{\pi }{8}\int_{0}^{\pi /2}\frac{2cos(x)sin(x)}{cos^{4}(x) + sin^{4}(x)}dx
I = \frac{\pi }{8}\int_{0}^{\pi /2}\frac{sin(2x)}{(cos^{2}(x))^{2} + (sin^{2}(x))^{2}}dx
sin^{2}x = t
sin2xdx = dt
x = 0\rightarrow t = 0
x = \frac{\pi }{2}\rightarrow t = 1
I = \frac{\pi }{8}\int_{0}^{1}\frac{1}{(1-t)^{2} + (t)^{2}}dt
I = \frac{\pi }{16}\int_{0}^{1}\frac{1}{(t-\frac{1}{2})^{2} + (\frac{1}{2})^{2}}dt
I = \frac{\pi }{16}.\frac{1}{\frac{1}{2}}.tan^{-1}(\frac{t-\frac{1}{2}}{\frac{1}{2}})_{0}^{1}
I = \frac{\pi }{8}.tan^{-1}(2t-1 )_{0}^{1}
I = \frac{\pi }{8}.[tan^{-1}(1)-tan^{-1}(-1)]
I = \frac{\pi }{4}.tan^{-1}(1)
I = \frac{\pi }{4}.\frac{\pi }{4} = \frac{\pi ^{2}}{16}

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