int_0^{pi / 4} frac{cos ^2 x}{cos ^2 x+4 sin | vedclass

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(D) Let $I = \int_0^{\pi / 4} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} dx$. Divide numerator and denominator by $\cos^2 x$: $I = \int_0^{\pi / 4} \frac{

Vedclass · Accepted Answer

$\frac{\pi}{12} + \frac{1}{3} 	an^{-1} 2$

Vedclass · Answer

(D) Let $I = \int_0^{\pi / 4} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} dx$. Divide numerator and denominator by $\cos^2 x$: $I = \int_0^{\pi / 4} \frac{1}{1 + 4 	an^2 x} dx$. Let $u = 	an x$,then $du = \sec^2 x dx = (1 + u^2) dx$,so $dx = \frac{du}{1 + u^2}$. When $x = 0, u = 0$. When $x = \pi / 4, u = 1$. $I = \int_0^1 \frac{1}{(1 + 4u^2)(1 + u^2)} du$. Using partial fractions: $\frac{1}{(1 + 4u^2)(1 + u^2)} = \frac{A}{1 + 4u^2} + \frac{B}{1 + u^2}$. $1 = A(1 + u^2) + B(1 + 4u^2)$. For $u^2 = -1, 1 = B(1 - 4) \implies B = -1/3$. For $u^2 = -1/4, 1 = A(1 - 1/4) \implies A = 4/3$. $I = \int_0^1 (\frac{4/3}{1 + 4u^2} - \frac{1/3}{1 + u^2}) du$. $I = \frac{4}{3} \int_0^1 \frac{1}{1 + (2u)^2} du - \frac{1}{3} \int_0^1 \frac{1}{1 + u^2} du$. $I = \frac{4}{3} [\frac{1}{2} 	an^{-1}(2u)]_0^1 - \frac{1}{3} [	an^{-1} u]_0^1$. $I = \frac{2}{3} 	an^{-1} 2 - \frac{1}{3} (\frac{\pi}{4}) = \frac{2}{3} 	an^{-1} 2 - \frac{\pi}{12}$.

$\int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=$

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