Question

How do you find the integral of $ \dfrac{1}{1+{{\sin }^{2}}x} $ ?

Answer 1

Hint: Here in this question, we have to integrate the given trigonometric expression. There are various integration formulae which we will be using here. Apart from that, Pythagoras identities of trigonometry will also be used. We will solve using the substitution method in integration. Some important formulae used are:
 $ \begin{align}
  & \Rightarrow \int{1dx}=x+C \\
 & \Rightarrow \int{{{\sec }^{2}}xdx}=\tan x+C \\
\end{align} $

Complete step by step answer:
Now, let’s solve the question.
We have to apply integration formulae to solve the trigonometric expression.
First, write the expression given in the question and denote it with ‘I’.
 $ \Rightarrow I=\int{\dfrac{1}{1+{{\sin }^{2}}x}}dx $
Now, divide numerator and denominator with $ {{\cos }^{2}}x $ , we get:
 $ \Rightarrow I=\int{\dfrac{\dfrac{1}{{{\cos }^{2}}x}}{\dfrac{1+{{\sin }^{2}}x}{{{\cos }^{2}}x}}}dx $
As we know some of the trigonometric identities and conversions. Let’s discuss them too.
 $ \Rightarrow $ sinx = $ \dfrac{1}{\cos ecx} $ or cosecx = $ \dfrac{1}{\sin x} $
 $ \Rightarrow $ cosx = $ \dfrac{1}{\sec x} $ or secx = $ \dfrac{1}{\cos x} $
 $ \Rightarrow $ tanx = $ \dfrac{1}{\cot x} $ or cotx = $ \dfrac{1}{\tan x} $
 $ \Rightarrow $ tan $ \theta $ = $ \dfrac{\sin \theta }{\cos \theta } $
 $ \Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x=1 $
 $ \Rightarrow 1+{{\tan }^{2}}x={{\sec }^{2}}x $
 $ \Rightarrow 1+{{\cot }^{2}}x=\cos e{{c}^{2}}x $
As we studied in the above formulae,
 $ \Rightarrow $ cosx = $ \dfrac{1}{\sec x} $ or secx = $ \dfrac{1}{\cos x} $
 $ \Rightarrow $ tan $ \theta $ = $ \dfrac{\sin \theta }{\cos \theta } $
Do all the necessary conversions and simplify:
 $ \Rightarrow I=\int{\dfrac{{{\sec }^{2}}x}{{{\sec }^{2}}x+{{\tan }^{2}}x}}dx $
We know that
 $ \Rightarrow 1+{{\tan }^{2}}x={{\sec }^{2}}x $
Put this value now:
 $ \Rightarrow I=\int{\dfrac{{{\sec }^{2}}x}{\left( 1+{{\tan }^{2}}x \right)+{{\tan }^{2}}x}}dx $
Open the brackets and solve:
 $ \Rightarrow I=\int{\dfrac{{{\sec }^{2}}x}{1+2{{\tan }^{2}}x}}dx $
Now we will use substitution method. Substitute tanx = $ \dfrac{1}{\sqrt{2}}\tan \theta $ .
So,
 $ \theta ={{\tan }^{-1}}\left( \sqrt{2}\tan x \right).....(i) $
 $ \begin{align}
  & \Rightarrow \int{\tan xdx=}\dfrac{1}{\sqrt{2}}\int{\tan \theta d\theta } \\
 & \Rightarrow {{\sec }^{2}}x=\dfrac{1}{\sqrt{2}}{{\sec }^{2}}\theta d\theta \\
\end{align} $
Now, we have to replace old values with all new values, we will get:
 $ \Rightarrow I=\int{\dfrac{\dfrac{1}{\sqrt{2}}{{\sec }^{2}}\theta }{1+2{{\left( \dfrac{1}{\sqrt{2}}\tan \theta \right)}^{2}}}d\theta } $
Next step is to take $ \dfrac{1}{\sqrt{2}} $ and open the brackets:
 $ \Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\dfrac{{{\sec }^{2}}\theta }{1+2\times \dfrac{1}{2}{{\tan }^{2}}\theta }d\theta } $
On solving further, we will get:
 $ \Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\dfrac{{{\sec }^{2}}\theta }{1+{{\tan }^{2}}\theta }d\theta } $
We will again use $ 1+{{\tan }^{2}}x={{\sec }^{2}}x $ in our expression.
 $ \Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\dfrac{{{\sec }^{2}}\theta }{{{\sec }^{2}}\theta }d\theta } $
Cancel the terms, we will get:
 $ \Rightarrow I=\dfrac{1}{\sqrt{2}}\int{1d\theta } $
As we know
 $ \Rightarrow \int{1dx}=x+C $
So,
 $ \Rightarrow I=\dfrac{1}{\sqrt{2}}\theta +C $
Put the equation(i) in it:
 $ \Rightarrow I=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \sqrt{2}\tan x \right)+C $
So this is the final answer.

Note:
 Though it’s too complicated to solve these types of questions, if you are familiar with trigonometric formulae and methods of integration then you will definitely find the solution. After integrating tanx = $ \dfrac{1}{\sqrt{2}}\tan \theta $ don’t forget to change the dx into $ d\theta $ . Otherwise, in the final step, you will not be able to substitute the value of $ \theta $, and marks will also be deducted.