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 :
$\int {\left[ {uv} \right] dx} = u\int v {\text{ dx}} - \int {\left[ {\left( {\dfrac{d}{{dx}}u} \right) \times \int v dx} \right] dx}$