Practicing Success
Practicing Success
CUET Preparation Today
\(\int \frac{\sin^2x}{1+\cos x}dx\)= |
\(x-\sin x+c\) \(x+\sin x+c\) \(x+\cos x+c\) \(x-\cos x+c\) |
\(x-\sin x+c\) |
\(\int \frac{\sin^2xdx}{1+\cos x}=\int\frac{1-\cos^2x}{1+\cos^2x}dx\) $=\int 1-\cos x dx$ $=x-\sin x+C$ |
