Question

What is $\int{\dfrac{x}{1+\sin x}dx}$?

Answer 1

Hint: We first multiply with $1-\sin x$ to simplify the trigonometric function of $\dfrac{x}{1+\sin x}$. We convert the trigonometric function in its integral form and then apply it by parts to find the solution of $\int{\dfrac{x}{1+\sin x}dx}$.

Complete step by step solution:
To complete the integration, we first simplify the function $\dfrac{x}{1+\sin x}$. We multiply both numerator and denominator with $1-\sin x$.
We get $\dfrac{x\left( 1-\sin x \right)}{\left( 1+\sin x \right)\left( 1-\sin x \right)}=\dfrac{x\left( 1-\sin x \right)}{1-{{\sin }^{2}}x}=\dfrac{x\left( 1-\sin x \right)}{{{\cos }^{2}}x}$.
Now we simplify the trigonometric part and get $\dfrac{x\left( 1-\sin x \right)}{{{\cos }^{2}}x}=x\left( {{\sec }^{2}}x-\tan x\sec x \right)$.
We now individually find the integration of the trigonometric part where we get
$\int{\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}=\tan x-\sec x$.
We now apply the by parts to find the integral of $\int{x\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}$.
Integration by parts method is usually used for the multiplication of the functions and their integration.
Let’s assume $f\left( x \right)=g\left( x \right)h\left( x \right)$. We need to find the integration of $\int{f\left( x \right)dx}=\int{g\left( x \right)h\left( x \right)dx}$.
We take $u=g\left( x \right),v=h\left( x \right)$. This gives $\int{f\left( x \right)dx}=\int{uvdx}$.
The theorem of integration by parts gives $\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx}$.
For our integration $\int{x\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}$, we take $u=x,v=\left( {{\sec }^{2}}x-\tan x\sec x \right)$.
Now we complete the integration
\[\int{x\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}=x\int{\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}-\int{\left( \dfrac{d\left( x \right)}{dx}\int{\left( {{\sec }^{2}}x-\tan x\sec x \right)dx} \right)dx}\].
The integration formula for $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$.
We apply these formulas to complete the integration and get
\[\int{x\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}=x\left( \tan x-\sec x \right)-\int{\left( \tan x-\sec x \right)dx}\].
We have one more integration part.
So, \[\int{x\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}=x\left( \tan x-\sec x \right)-\log \left| \sec x \right|+\log \left| \sec x+\tan x \right|+c\].
Here $c$ is the integral constant.
We simplify the form to get
\[\begin{align}
  & x\left( \tan x-\sec x \right)-\log \left| \sec x \right|+\log \left| \sec x+\tan x \right| \\
 & =x\left( \tan x-\sec x \right)+\log \left| \dfrac{\sec x+\tan x}{\sec x} \right| \\
 & =x\left( \tan x-\sec x \right)+\log \left| 1+\sin x \right| \\
\end{align}\]
The integral form of $\int{\dfrac{x}{1+\sin x}dx}$ is \[x\left( \tan x-\sec x \right)+\log \left| 1+\sin x \right|+c\].

Note:
In case one of two functions are missing and we need to form the by parts method, we will take the multiplying constant 1 as the second function.
For example: if we need to find \[\int{\ln xdx}\], we have only one function. So, we take constant 1 as the second function where $u=\ln x,v=1$. But we need to remember that we won’t perform by parts by taking $u=1$.