Question

How do you find the integral of $ \dfrac{1}{{{\sin }^{2}}x}+\cos \left( 2x \right) $ ?

Answer 1

Hint:
We will use the standard integral formula to solve the above question. We will use the integral formula of the trigonometric terms such as $ \int{\cos dx=\sin x+C} $ , $ \int{\sin xdx=-\cos x+C} $ , also $ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C} $ . Also, we will use substitution method to solve the above equation.

Complete step by step answer:
We will use the concept of integration to solve the above question.
Since , we have to find the integral of $ \dfrac{1}{{{\sin }^{2}}x}+\cos \left( 2x \right) $ . So, we will write it in mathematical form as: $ \int{\left( \dfrac{1}{{{\sin }^{2}}x}+\cos \left( 2x \right) \right)}dx $
Since, we know that $ \int{\left( A+B \right)dx}=\int{Adx+\int{Bdx}} $ .
So, we can write $ \int{\left( \dfrac{1}{{{\sin }^{2}}x}+\cos \left( 2x \right) \right)}dx $ = $ \int{\dfrac{1}{{{\sin }^{2}}x}}dx+\int{\cos \left( 2x \right)dx} $
Now, let us call the left part of the integral as Integral 1 and second right part as Integral 2.
So, let us first solve Integral 1.
Integral 1 = $ \int{\dfrac{1}{{{\sin }^{2}}x}}dx $
Now, we will divide both numerator and denominator by $ {{\cos }^{2}}x $ . Then, we can write the above integral as:
Integral 1 = $ \int{\dfrac{{{\sec }^{2}}x}{{{\tan }^{2}}x}}dx $
Let us assume $ t=\tan x $
Then, $ \dfrac{dt}{dx}={{\sec }^{2}}x $ , so $ dt={{\sec }^{2}}xdx $
 $ \Rightarrow \int{\dfrac{{{\sec }^{2}}x}{{{\tan }^{2}}x}}dx=\int{\dfrac{1}{{{t}^{2}}}dt} $
And, we know that $ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C} $ .
 $ \Rightarrow \int{\dfrac{1}{{{t}^{2}}}dt}=\int{{{t}^{-2}}dt}=\dfrac{{{t}^{-1}}}{-1}+C $
 $ \Rightarrow \int{\dfrac{1}{{{\sin }^{2}}x}}dx=\dfrac{{{t}^{-1}}}{-1}+C $ , where $ t=\tan x $
Hence, $ \int{\dfrac{1}{{{\sin }^{2}}x}}dx=\dfrac{{{\left( \tan x \right)}^{-1}}}{-1}+C $ .
Since, we know that $ \dfrac{1}{\tan x}=\cot x $ , so $ \int{\dfrac{1}{{{\sin }^{2}}x}}dx=-\cot x+C $
Now, we will solve Integral 2.
Integral 2 = \[\int{\cos \left( 2x \right)}dx\]
Let us assume that 2x = u.
So, $ \dfrac{du}{dx}=2 $
 $ \Rightarrow dx=\dfrac{du}{2} $
So, after substituting 2x = u and $ dx=\dfrac{du}{2} $ in Integral 2, we will get:
 $ \Rightarrow \int{\cos 2xdx=}\int{\dfrac{\cos u}{2}du} $
 $ =\dfrac{1}{2}\int{\cos udu} $
We know that $ \int{\cos dx=\sin x+C} $
 $ \Rightarrow \dfrac{1}{2}\int{\cos udu}=\dfrac{\sin u}{2}+C $
Hence, \[\int{\cos \left( 2x \right)}dx=\dfrac{\sin u}{2}+C\], where u = 2x.
So, Integral 2 = \[\int{\cos \left( 2x \right)}dx=\dfrac{\sin 2x}{2}+C\]
Now, we will add both the Integral 1 and Integral 2 to get $ \int{\left( \dfrac{1}{{{\sin }^{2}}x}+\cos \left( 2x \right) \right)}dx $ .
 $ \Rightarrow \int{\left( \dfrac{1}{{{\sin }^{2}}x}+\cos \left( 2x \right) \right)}dx=Integral1+Integral2 $
 $ \Rightarrow \int{\left( \dfrac{1}{{{\sin }^{2}}x}+\cos \left( 2x \right) \right)}dx=-\cot x+C+\dfrac{\sin 2x}{2}+C=-\cot x+\dfrac{\sin 2x}{2}+C $
This is our required solution.

Note:
 Students are required to memorize all the standard integration formulas and also standard derivatives formula otherwise they will not be able to solve the question as they can see we have also used the derivatives formula while solving the Integral 1 part in the above question along with the standard integral formula.