Question

Find the value of \[\int {\left[ {\dfrac{{\left( {x + sin{\text{ }}x} \right)}}{{\left( {1 + cos{\text{ }}x} \right)}}} \right] } {\text{ }}dx\]

Answer 1

Hint: We have to integrate \[{\text{ }}\dfrac{{\left( {x + sin{\text{ }}x} \right)}}{{\left( {1 + cos{\text{ }}x} \right)}}\] with respect to \[x\] . We solve this using integration of by parts and using various formulas of trigonometric functions . We firstly apply the double angle formula in cos function then we solve the integration by splitting it into parts and after applying by-parts we get the solution of the integral.

Complete step-by-step answer:
Given : \[\int {\left[ {\dfrac{{\left( {x + sin{\text{ }}x} \right)}}{{\left( {1 + cos{\text{ }}x} \right)}}} \right] } {\text{ }}dx\]
Let \[I = \int {\left[ {\dfrac{{\left( {x + sin{\text{ }}x} \right)}}{{\left( {1 + cos{\text{ }}x} \right)}}} \right] } {\text{ }}dx\]
We have to integrate \[I\] with respect to \[x\]
As we know , $cos2x = 2co{s^2}x - 1$
Using this formula , we get
 \[I = \int {\left[ {\dfrac{{\left( {x + sin{\text{ }}x} \right)}}{{\left( {1 + 2co{s^2}\left( {\dfrac{x}{2}} \right) - 1} \right)}}} \right] } {\text{ }}dx\]
 On simplifying , we get
 \[I = \int {\left[ {\dfrac{{\left( {x + sin{\text{ }}x} \right)}}{{2co{s^2}\left( {\dfrac{x}{2}} \right)}}} \right] } {\text{ }}dx\]
Now dividing numerator by \[2co{s^2}\left( {\dfrac{x}{2}} \right)\] and writing the terms separately , we get
 \[\left( {\cos x = \dfrac{1}{{\sec x}}} \right)\]
 \[I = \int {\left[ {\dfrac{x}{2}{{\sec }^2}\dfrac{x}{2} + \dfrac{1}{2}\sin x \times {{\sec }^2}\dfrac{x}{2}} \right] dx} \]
Also , \[\sin 2x = 2\sin x\cos x\]
Using value sin double angle , we get
 \[I = \int {\left[ {\dfrac{x}{2}{{\sec }^2}\dfrac{x}{2} + \dfrac{1}{2} \times 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \times {{\sec }^2}\dfrac{x}{2}} \right] dx} \]
  \[\left( {\cos x = \dfrac{1}{{\sec x}}} \right)\]
  \[\left( {\tan x = \dfrac{{\sin x}}{{\cos x}}} \right)\]
After simplifying the terms , we get
 \[I = \int {\left[ {\dfrac{x}{2}{{\sec }^2}\dfrac{x}{2} + \tan \dfrac{x}{2}} \right] dx} \]
Now , using formula of by - parts :
 \[\int {\left[ {uv} \right] dx} = u\int v {\text{ dx}} - \int {\left[ {\left( {\dfrac{d}{{dx}}u} \right) \times \int v dx} \right] dx} \]
We get,
 \[I = \dfrac{x}{2}\int {{{\sec }^2}\dfrac{x}{2}dx} - \dfrac{1}{2}\int {\left[ {\left( {\dfrac{d}{{dx}}x} \right) \times \int {{{\sec }^2}\dfrac{x}{2}} dx} \right] dx} + \int {\tan \dfrac{x}{2}dx} \]
Using integral formula \[\int {{{\sec }^2}x} dx = \tan x + c\] and [ derivative of${x^n} = n{x^{n - 1}}$] , we get
 \[I = \dfrac{x}{2}\tan \dfrac{x}{2} \times 2 + a - \dfrac{1}{2}\int {\left[ {1 \times \tan \dfrac{x}{2} \times 2} \right] dx} + \int {\tan \dfrac{x}{2}dx} \]
 \[I = \dfrac{x}{2}\tan \dfrac{x}{2} \times 2 + a - \int {\left[ {\tan \dfrac{x}{2}} \right] dx} + \int {\tan \dfrac{x}{2}dx} \]
After cancelling terms , we get
 \[I = x\tan \dfrac{x}{2} + a\]
Where \[a\] is integration constant
So, the correct answer is “ \[I = x\tan \dfrac{x}{2} + a\] ”.

Note: As the question was of indefinite integral that’s why we added integration constant. If the question would be of definite integral then we don’t add integral constant to the final answer .
We use the formula of By-Parts to integrate two functions of a single variable $x$by taking one functions as $u$ and second function as $v$ and then applying the formula :
 \[\int {\left[ {uv} \right] dx} = u\int v {\text{ dx}} - \int {\left[ {\left( {\dfrac{d}{{dx}}u} \right) \times \int v dx} \right] dx} \]