Question:
Evaluate the following integrals:
$\int \frac{1-\sin 2 x}{x+\cos ^{2} x} d x$
Solution:
Assume $x+\cos ^{2} x=t$
$d\left(x+\cos ^{2} x\right)=d t$
$(1+(-2 \cos x \cdot \sin x)) d x=d t$
$2 \sin x \cdot \cos x=\sin 2 x$
$(1-\sin 2 x) d x=d t$\
Put $t$ and $d t$ in given equation we get
$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t}}$
$=\ln |\mathrm{t}|+\mathrm{c}$
But $t=x+\cos ^{2} x$
$=\ln \left|x+\cos ^{2} x\right|+c$
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