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CUET
-- Mathematics - Section B1
Indefinite Integration
\(\int \frac{\sin^2x}{1+\cos x}dx\)=
\(x-\sin x+c\)
\(x+\sin x+c\)
\(x+\cos x+c\)
\(x-\cos x+c\)
\(\int \frac{\sin^2xdx}{1+\cos x}=\int\frac{1-\cos^2x}{1+\cos^2x}dx\)
$=\int 1-\cos x dx$
$=x-\sin x+C$
