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\( \int \frac{\sin 2 x}{\left(\sin ^{2} x+2 \cos ^{2} x\right.} d x= \)
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Verified Answer
The correct answer is:
\( -\log \left(1+\cos ^{2} x\right)+C \)
Given that,$I=\int \frac{\sin 2 x}{\sin ^{2} x+2 \cos ^{2} x} d x$
Since, $\sin ^{2} x=1-\cos ^{2} x .$ So,
$=\int \frac{\sin 2 x d x}{1+\cos ^{2} x}$
Let $1+\cos ^{2} x=t .$ So,
$-2 \cos x \sin x d x=d t$
$\Rightarrow \sin 2 x d x=-d t$
$\Rightarrow-\int \frac{d t}{t}=-\log t+c=-\log \left(1+\cos ^{2} x\right)+c$
Since, $\sin ^{2} x=1-\cos ^{2} x .$ So,
$=\int \frac{\sin 2 x d x}{1+\cos ^{2} x}$
Let $1+\cos ^{2} x=t .$ So,
$-2 \cos x \sin x d x=d t$
$\Rightarrow \sin 2 x d x=-d t$
$\Rightarrow-\int \frac{d t}{t}=-\log t+c=-\log \left(1+\cos ^{2} x\right)+c$
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