Question

$\int \frac{se{c}^2\left(x\right)}{4ta{n}^2\left(x\right)+9}dx$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Here, if we put tan(x) = t the numerator $se{c}^2\left(x\right)$ dx will become dt. As $se{c}^2\left(x\right)$ is the derivative of tan(x). And we’ll be left with a quadratic equation in the denominator which we can solve.

Let’s substitute tan(x) = t

So, $se{c}^2\left(x\right)$ dx = dt

And the given integral would be like

$\int \frac{1}{4t^2+9}dtOr\frac{1}{4}\int \frac{1}{t^2+\frac{9}{4}}dtOr\frac{1}{4}\int \frac{1}{t^2+3{\left(\frac{3}{2}\right)}^2}dt$

We can see that this is of the form $\int \frac{1}{x^2+a^2}dx$

After using the corresponding formula and substituting back the value of “ t “ which is tan(x) we get the final answer equal to

$\frac{1}{6}ta{n}^{-1}\left(\frac{2tan\left(x\right)}{3}\right)+C$