Practicing Success
Practicing Success
CUET Preparation Today
| $\int \frac{1-\tan^2 x}{1+\tan^2 x}dx$? |
$\frac{\sin 2x}{2}+C$ $\frac{\cos 2x}{2}+C$ $\frac{\sin 2x}{4}+C$ $\sin 2x+C$ |
| $\frac{\sin 2x}{2}+C$ |
| $\int \frac{1-\tan^2 x}{1+\tan^2 x}dx$=$\int \frac{1-\tan^2 x}{\sec^2 x}dx=\int (\cos^2x-\sin^2x)dx=\int \cos 2xdx=\frac{sin 2x}{2}+C$ |
