Question

$\int \frac{1}{4{\sin }^{2}x+9{\cos }^{2}x}dx$ will be equal to -

• A. $\frac{1}{3}{\tan }^{-1}\left(\frac{2\tan \left(x\right)}{3}\right)+C$
• B. $\frac{1}{6}{\tan }^{-1}\left(\frac{2\tan \left(x\right)}{3}\right)+C$
• C. $\frac{1}{6}{\tan }^{-1}\left(\frac{2\tan \left(x\right)}{6}\right)+C$
• D. $\frac{1}{6}{\tan }^{-1}\left(\frac{2\tan \left(x\right)}{5}\right)+C$

Answer 1

The correct option is B $\frac{1}{6}{\tan }^{-1}\left(\frac{2\tan \left(x\right)}{3}\right)+C$

In solving the integration problem it is important to see the form in which integrand is. Here, the given integrand is in $\frac{1}{a{\cos }^{2}x+b{\sin }^{2}x}$.
To solve such integrals we need to multiply numerator as well as denominator of the integrand by ${\sec }^{2}x$.
Let’s do that and see what we get.
$\int \frac{{\sec }^{2}xdx}{4{\tan }^{2}x+9}$
Now, we know how to solve this. We have already solved these forms.
Here, if we put tan(x) = t the numerator ${\sec }^2\left(x\right)dx$ will become dt, as ${\sec }^2\left(x\right)$ 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, ${\sec }^2\left(x\right)dx=dt$
And the given integral would be like
$\int \frac{1}{4t^2+9dt}$
$\Rightarrow \frac{1}{4}\int \frac{1}{t^2+\frac{9}{4}}dt$
$\Rightarrow \frac{1}{4}\int \frac{1}{t^2+{\left(\frac{3}{2}\right)}^2}dt$
We can see that this is of the form $\int \frac{1}{x^2+a^2}dx$
So, we’ll use the corresponding formula which is
$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$
So,we get $\frac{1}{6}{\tan }^{-1}\left(\frac{2t}{3}\right)+C$
Let’s substitute value of "t" which is tan(x).
So, the final answer would be$=\frac{1}{6}{\tan }^{-1}\left(\frac{2\tan \left(x\right)}{3}\right)+C$