The value of the definite integral int limits | vedclass

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Question

(A) Let $I = \int \limits_0^{\pi / 2} \frac{\sin x \cos x}{1+\cos ^4 x} d x$.Substitute $\cos ^2 x = t$. Then,differentiating both sides with respect

Vedclass · Accepted Answer

$\frac{\pi}{8}$

Vedclass · Answer

(A) Let $I = \int \limits_0^{\pi / 2} \frac{\sin x \cos x}{1+\cos ^4 x} d x$.Substitute $\cos ^2 x = t$. Then,differentiating both sides with respect to $x$,we get $2 \cos x (-\sin x) d x = d t$,which implies $\sin x \cos x d x = -\frac{1}{2} d t$.Change the limits of integration:When $x = 0$,$t = \cos ^2(0) = 1$.When $x = \frac{\pi}{2}$,$t = \cos ^2(\frac{\pi}{2}) = 0$.Substituting these into the integral:$I = \int \limits_1^0 \frac{-1/2}{1+t^2} d t = \frac{1}{2} \int \limits_0^1 \frac{1}{1+t^2} d t$.Evaluating the integral:$I = \frac{1}{2} [	an ^{-1}(t)]_0^1 = \frac{1}{2} (	an ^{-1}(1) - 	an ^{-1}(0)) = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}$.

The value of the definite integral $\int \limits_0^{\pi / 2} \frac{\sin x \cos x}{1+\cos ^4 x} d x$ is:

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