int frac{sin x cos x}{a cos^2 x + b sin^2 x} | vedclass

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(A) Let $I = \int \frac{\sin x \cos x}{a \cos^2 x + b \sin^2 x} dx$. Substitute $t = a \cos^2 x + b \sin^2 x$. Then $dt = (a \cdot 2 \cos x(-\sin x) +

Vedclass · Accepted Answer

$\frac{1}{2(b - a)} \log |a \cos^2 x + b \sin^2 x| + C$

Vedclass · Answer

(A) Let $I = \int \frac{\sin x \cos x}{a \cos^2 x + b \sin^2 x} dx$. Substitute $t = a \cos^2 x + b \sin^2 x$. Then $dt = (a \cdot 2 \cos x(-\sin x) + b \cdot 2 \sin x \cos x) dx = 2(b - a) \sin x \cos x dx$. So,$\sin x \cos x dx = \frac{dt}{2(b - a)}$. Substituting these into the integral,we get: $I = \int \frac{1}{t} \cdot \frac{dt}{2(b - a)} = \frac{1}{2(b - a)} \int \frac{1}{t} dt$. $I = \frac{1}{2(b - a)} \log |t| + C$. Substituting back $t = a \cos^2 x + b \sin^2 x$,we get: $I = \frac{1}{2(b - a)} \log |a \cos^2 x + b \sin^2 x| + C$.

$\int \frac{\sin x \cos x}{a \cos^2 x + b \sin^2 x} dx = $

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