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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int_0^{\frac{\pi}{4}} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int_0^{\frac{\pi}{4}} \frac{1}{1 + 4	an^2 x} dx$. Let $u = 	an x$,then $du = \sec^2 x dx$. Since $\sec^2 x = 1 + 	an^2 x = 1 + u^2$,we have $dx = \frac{du}{1+u^2}$. When $x = 0, u = 0$. When $x = \frac{\pi}{4}, u = 1$. $I = \int_0^1 \frac{1}{(1+4u^2)(1+u^2)} du$. 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 $u^2 = -1, 1 = B(1-4) \Rightarrow B = -\frac{1}{3}$. For $u^2 = -\frac{1}{4}, 1 = A(1-\frac{1}{4}) \Rightarrow A = \frac{4}{3}$. $I = \int_0^1 (\frac{4/3}{1+4u^2} - \frac{1/3}{1+u^2}) du = [\frac{4}{3} \cdot \frac{1}{2} 	an^{-1}(2u) - \frac{1}{3} 	an^{-1}(u)]_0^1$. $I = [\frac{2}{3} 	an^{-1}(2u) - \frac{1}{3} 	an^{-1}(u)]_0^1 = \frac{2}{3} 	an^{-1}(2) - \frac{1}{3} 	an^{-1}(1) = \frac{2}{3} 	an^{-1}(2) - \frac{1}{3} \cdot \frac{\pi}{4} = \frac{2}{3} 	an^{-1}(2) - \frac{\pi}{12}$.

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

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