int_0^{pi / 4} frac{sec x}{3 cos x+4 sin x} d | vedclass

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Question

(B) Let $I = \int_0^{\pi / 4} \frac{\sec x}{3 \cos x+4 \sin x} d x$. Multiply the numerator and denominator by $\sec x$: $I = \int_0^{\pi / 4} \frac{\

Vedclass · Accepted Answer

$\frac{1}{4} \log \left(\frac{7}{3}\right)$

Vedclass · Answer

(B) Let $I = \int_0^{\pi / 4} \frac{\sec x}{3 \cos x+4 \sin x} d x$. Multiply the numerator and denominator by $\sec x$: $I = \int_0^{\pi / 4} \frac{\sec^2 x}{3 + 4 	an x} d x$. Let $u = 3 + 4 	an x$. Then $du = 4 \sec^2 x d x$, so $\sec^2 x d x = \frac{du}{4}$. When $x = 0$, $u = 3 + 4(0) = 3$. When $x = \pi / 4$, $u = 3 + 4(1) = 7$. Substituting these into the integral: $I = \int_3^7 \frac{1}{u} \cdot \frac{du}{4} = \frac{1}{4} [\log |u|]_3^7$. $I = \frac{1}{4} (\log 7 - \log 3) = \frac{1}{4} \log \left(\frac{7}{3}ight)$.

$\int_0^{\pi / 4} \frac{\sec x}{3 \cos x+4 \sin x} d x=$

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