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) $2 \sec^2 x + c$

✨ Free AI Tools for You

Powered by Chatterbot AI

Answer


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

Given: $$ \frac{d \tan x}{dx} = \sec^2 x $$

We know that: $$ \int \sec^2 x , dx = \tan x $$

Now, consider the integral: $$ \int \frac{2}{1-\sin^2 x} , dx $$

Since $1 - \sin^2 x = \cos^2 x$, we can rewrite the integral as: $$ \int \frac{2}{\cos^2 x} , dx = \int 2 \sec^2 x , dx $$

Evaluating the integral: $$ \int 2 \sec^2 x , dx = 2 \int \sec^2 x , dx = 2 \tan x + c $$

Thus, the solution is: $$ 2 \tan x + c $$


Was this helpful?

India's 1st AI Doubt Solver for CBSE, JEE, and NEET

Ask a Question for Free

and then it's just ₹212 a month