$\int \frac{d x}{\cos 2 x+\sin ^{2} x}=$
$\int \frac{d x}{\cos 2 x+\sin ^{2} x}=$
- $\sin x+c$
- $\tan x+c$
- $\sec ^{2} x+c$
- $\cos x+c$
Solution
$\begin{aligned} I &=\int \frac{d x}{1-2 \sin ^{2} x+\sin ^{2} x} \\ &=\int \frac{d x}{1-\sin ^{2} x}=\int \frac{d x}{\cos ^{2} x}=\int \sec ^{2} x d x=\tan x+c \end{aligned}$
Asked in: MHT CET 2020 (15 Oct Shift 1)
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