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

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(C) Let $I = \int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x$. Dividing numerator and denominator by $\cos^2 x$, we get: $I = \int_0^{\pi

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x$. Dividing numerator and denominator by $\cos^2 x$, we get: $I = \int_0^{\pi / 4} \frac{1}{1+4 	an ^2 x} d x$. Let $	an x = t$, then $\sec^2 x dx = dt$, which implies $dx = \frac{dt}{1+t^2}$. When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$. So, $I = \int_0^1 \frac{1}{(1+t^2)(1+4t^2)} dt$. Using partial fractions: $\frac{1}{(1+t^2)(1+4t^2)} = \frac{1}{3} \left( \frac{4}{1+4t^2} - \frac{1}{1+t^2} ight)$. $I = \frac{1}{3} \int_0^1 \left( \frac{4}{1+4t^2} - \frac{1}{1+t^2} ight) dt$. $I = \frac{1}{3} \left[ 2 	an^{-1}(2t) - 	an^{-1}(t) ight]_0^1$. $I = \frac{1}{3} \left[ (2 	an^{-1}(2) - 	an^{-1}(1)) - (0 - 0) ight]$. $I = \frac{1}{3} \left[ 2 	an^{-1}(2) - \frac{\pi}{4} ight] = \frac{2}{3} 	an^{-1}(2) - \frac{\pi}{12}$.

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

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