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Question: What is \(\int{\dfrac{x}{1+\sin x}dx}\)?...

What is ∫x1+sin⁡xdx\int{\dfrac{x}{1+\sin x}dx}?

Explanation

Solution

We first multiply with 1−sin⁡x1-\sin x to simplify the trigonometric function of x1+sin⁡x\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 ∫x1+sin⁡xdx\int{\dfrac{x}{1+\sin x}dx}.

Complete step by step solution:
To complete the integration, we first simplify the function x1+sin⁡x\dfrac{x}{1+\sin x}. We multiply both numerator and denominator with 1−sin⁡x1-\sin x.
We get x(1−sin⁡x)(1+sin⁡x)(1−sin⁡x)=x(1−sin⁡x)1−sin⁡2x=x(1−sin⁡x)cos⁡2x\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 x(1−sin⁡x)cos⁡2x=x(sec⁡2x−tan⁡xsec⁡x)\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
∫(sec⁡2x−tan⁡xsec⁡x)dx=tan⁡x−sec⁡x\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 ∫x(sec⁡2x−tan⁡xsec⁡x)dx\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(x)=g(x)h(x)f\left( x \right)=g\left( x \right)h\left( x \right). We need to find the integration of ∫f(x)dx=∫g(x)h(x)dx\int{f\left( x \right)dx}=\int{g\left( x \right)h\left( x \right)dx}.
We take u=g(x),v=h(x)u=g\left( x \right),v=h\left( x \right). This gives ∫f(x)dx=∫uvdx\int{f\left( x \right)dx}=\int{uvdx}.
The theorem of integration by parts gives ∫uvdx=u∫vdx−∫(dudx∫vdx)dx\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx}.
For our integration ∫x(sec⁡2x−tan⁡xsec⁡x)dx\int{x\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}, we take u=x,v=(sec⁡2x−tan⁡xsec⁡x)u=x,v=\left( {{\sec }^{2}}x-\tan x\sec x \right).
Now we complete the integration
∫x(sec⁡2x−tan⁡xsec⁡x)dx=x∫(sec⁡2x−tan⁡xsec⁡x)dx−∫(d(x)dx∫(sec⁡2x−tan⁡xsec⁡x)dx)dx\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 ∫xndx=xn+1n+1+c\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c.
We apply these formulas to complete the integration and get
∫x(sec⁡2x−tan⁡xsec⁡x)dx=x(tan⁡x−sec⁡x)−∫(tan⁡x−sec⁡x)dx\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, ∫x(sec⁡2x−tan⁡xsec⁡x)dx=x(tan⁡x−sec⁡x)−log⁡∣sec⁡x∣+log⁡∣sec⁡x+tan⁡x∣+c\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 cc is the integral constant.
We simplify the form to get

& 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{aligned}$$ **The integral form of $\int{\dfrac{x}{1+\sin x}dx}$ is $$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 $u=1$.