Integral problem: [int] arcsin ( 2x / ( 1+x^2 ) ) dx

[grapz]

I am having trouble doing this hard integral problem

[int] arcsin ( 2x / ( 1+x^2 ) ) dx

I think i have to somehow get the value inside the arcsin to contain a sin so it cancels out and makes this integral easier, but i am unsure on how to do this.

 

[galactus]

Re: Integral problem

Let's use integration by parts.

\(\displaystyle \int{sin^{-1}(\frac{2x}{1+x^{2}})}dx\)

Let \(\displaystyle u=sin^{-1}(\frac{2x}{1+x^{2}}), \;\ dv=dx, \;\ du=\frac{-2}{1+x^{2}}dx, \;\ v=x\)

We get:

\(\displaystyle xsin^{-1}(\frac{2x}{1+x^{2}})+2\int{\frac{x}{1+x^{2}}}dx\)

Can you finish?.

 

[soroban]

Re: Integral problem

Hello, grapz!

This is a messy one . . .

\(\displaystyle \int \arcsin\left(\frac{2x}{1+x^2}\right)\,dx\)

I'm going to change the inverse trig function . . .

\(\displaystyle \text{Let: }\:\theta \:=\:\arcsin\left(\frac{2x}{1+x^2}\right) \quad\Rightarrow\quad \sin\theta \:=\:\frac{2x}{1+x^2} \:=\:\frac{opp}{hyp}\)

\(\displaystyle \theta\text{ is in a right triangle with: }\pp \,= \,2x,\;hyp \,= \,1+x^2\)

\(\displaystyle \text{Hence: }\:adj \,=\,1-x^2\quad\hdots\;\;\text{and: }\:\tan\theta \:= \:\frac{2x}{1-x^2} \quad\Rightarrow\quad\theta\;=\;\arctan\left(\frac{2x}{1-x^2}\right)\)

\(\displaystyle \text{The integral becomes: }\;\int\arctan\left(\frac{2x}{1-x^2}\right)dx\)

Integrate by parts . . .

. . ** .\(\displaystyle \begin{array}{ccccccc}u &=&\arctan\left(\dfrac{2x}{1-x^2}\right) & & dv &=& dx \\ du &=&\dfrac{2\,dx}{1+x^2} & & v &=& x \end{array}\)

\(\displaystyle \text{Then we have: }\;x\cdot\arctan\left(\frac{2x}{1-x^2}\right) - \int\frac{2x}{1+x^2}\,dx\)

. . . . . . \(\displaystyle =\;\boxed{x\cdot\arctan\left(\frac{2x}{1-x^2}\right) - \ln(1 + x^2) + C}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

\(\displaystyle \text{We have: }\;u \;=\;\arctan\left(\frac{2x}{1-x^2}\right)\)

\(\displaystyle \text{Differentiate:}\)

\(\displaystyle \frac{du}{dx} \;=\;\frac{1}{1 + \left(\frac{2x}{1-x^2}\right)^2} \cdot\frac{(1-x^2)\cdot2 - (2x)(-2x)}{(1-x^2)^2} \;= \;\frac{(1-x^2)^2}{(1-x^2)^2 + 4x^2}\cdot\frac{2-2x^2+4x^2}{(1-x^2)^2}\)

. . . \(\displaystyle =\;\frac{1}{1 - 2x^2 + x^4 + 4x^2}\cdot\frac{2+ 2x^2}{1} \;=\;\frac{1}{1+2x^2+x^4}\cdot\frac{2(1+x^2)}{1}\)

. . \(\displaystyle = \;\frac{1}{(1+x^2)^2}\cdot\frac{2(1+x^2)}{1} \;=\;\frac{2}{1+x^2}\quad\hdots\quad \text{Isn't that amazing?}\)

 

[grapz]

Re: Integral problem

Thanks for the replies.

I found a way to write the inner term in terms of sin. I used x = tan theta and the inner term just simplifies to sin2Q. So arcsin(sin2Q) is just 2Q.