CUET Preparation Today
The value of $\int\frac{\sin x\cos x}{\sqrt{1-\sin^4x}}$ is equal to: |
$\frac{1}{2}\sin^{-1}(\sin x)+C$ $-\frac{1}{2}\cos^{-1}(\sin^2 x)+C$ $\tan^{-1}(\sin^2 x)+C$ $\cot^{-1}(\sin x)+C$ |
$-\frac{1}{2}\cos^{-1}(\sin^2 x)+C$ |
Put $\sin^2 x = t⇒2\sin x.\cos x\,dx=dt$ $⇒I=\frac{1}{2}\int\frac{dt}{\sqrt{1-t^2}}=\frac{1}{2}\sin^{-1}(t)+C=\frac{1}{2}\sin^{-1}(\sin^2x)+C$ |
