Question

What is $ \int{\dfrac{{{\tan }^{2}}x}{\sec x}dx} $ ?

Answer 1

Hint: We first use the trigonometric form of $ {{\tan }^{2}}x={{\sec }^{2}}x-1 $ to simplify the fraction form of $ \dfrac{{{\tan }^{2}}x}{\sec x} $ . Then we break the integration in two parts and use the integral form of $ \int{\cos xdx}=\sin x $ and $ \int{\sec xdx}=\log \left| \sec x+\tan x \right| $ . We use an integral constant to find the final solution.

Complete step by step solution:
We first need to convert the numerator of the fraction $ \dfrac{{{\tan }^{2}}x}{\sec x} $ into the form of $ {{\tan }^{2}}x={{\sec }^{2}}x-1 $ .
We now simplify the given fraction in the form of
 $ \dfrac{{{\tan }^{2}}x}{\sec x}=\dfrac{{{\sec }^{2}}x-1}{\sec x}=\sec x-\dfrac{1}{\sec x} $ .
We know that
 $ \dfrac{1}{\sec x}=\cos x $ which gives $ \dfrac{{{\tan }^{2}}x}{\sec x}=\sec x-\cos x $ .
Now we integrate the function separately and get
 $ \int{\dfrac{{{\tan }^{2}}x}{\sec x}dx}=\int{\left( \sec x-\cos x \right)dx} $ .
The function gets separated and gives
 $ \int{\left( \sec x-\cos x \right)dx}=\int{\sec xdx}-\int{\cos xdx} $ .
We know that $ \int{\sec xdx}=\log \left| \sec x+\tan x \right| $ and $ \int{\cos xdx}=\sin x $ .
Therefore, the integral form is
 $ \int{\left( \sec x-\cos x \right)dx}=\log \left| \sec x+\tan x \right|-\sin x+c $ .
Here $ c $ is the integral constant.
Therefore, the solution of $ \int{\dfrac{{{\tan }^{2}}x}{\sec x}dx} $ is $ \log \left| \sec x+\tan x \right|-\sin x+c $ .
So, the correct answer is “ $ \log \left| \sec x+\tan x \right|-\sin x+c $ .”.

Note: We can also solve the integration using the base change for the variable $ z=\tan x $ . In that case the sum gets complicated but the final solution would be the same. It is better to watch out for the transformations in the trigonometric forms in the fractions and take that as the change in the variable.