Question

If $\frac{d\tan x}{dx}={\sec }^{2}x$, then $\int \frac{2}{1-{\sin }^{2}x}dx$ is equal to:

• A. $2\tan x+c$
• B. $2{\cos }^{2}x+c$
• C. ${\tan }^{2}x+c$
• D. $2{\sec }^{2}x+c$

Answer 1

The correct option is A $2\tan x+c$

If $\frac{d\tan x}{dx}={\sec }^{2}x$,
Then, $\int {\sec }^{2}xdx=\tan x$.

$\int \frac{2}{1-{\sin }^{2}x}dx=\int \frac{2}{{\cos }^{2}x}dx$
$=\int 2{\sec }^{2}xdx=2\tan x+c$