# Inverse Sine Integral

• Feb 26th 2006, 10:06 AM
alysha27
Inverse Sine Integral
Hi, I tried this problem and got an answer but I am not sure if it is the correct solution. If someone could check my solution, that'd be great.

Q: the integral of the inverse sine of 2x dx (ie: of sin^-1(2x) )

I used integration by parts and also substituiton and this is the answer I ended up with:

xsin^-1(2x) + 1/4*ln(|1-4x^2|) + C
• Feb 26th 2006, 10:29 AM
ThePerfectHacker
Quote:

Originally Posted by alysha27
Hi, I tried this problem and got an answer but I am not sure if it is the correct solution. If someone could check my solution, that'd be great.

Q: the integral of the inverse sine of 2x dx (ie: of sin^-1(2x) )

I used integration by parts and also substituiton and this is the answer I ended up with:

xsin^-1(2x) + 1/4*ln(|1-4x^2|) + C

You are almost right you misused the natural log,
Given,
$\displaystyle \int\sin^{-1}(2x)dx$ use the substitution $\displaystyle u=2x$ then by the substitution rule we have, $\displaystyle \frac{1}{2}\int\sin^{-1}udu$
Now consider this integral as,
$\displaystyle \int(1)\sin^{-1}udu$. Use integration by parts to get,
$\displaystyle u\sin^{-1}u-\int\frac{u}{\sqrt{1-u^2}}du$.
Notice this new integral is solvable by substitution, call $\displaystyle v=1-u^2$, thus,
$\displaystyle u\sin^{-1}u+\frac{1}{2}\int v^{-1/2}dv$.
Thus, by the exponent rule,
$\displaystyle u\sin^{-1}u+v^{1/2}$ substitute back,
$\displaystyle u\sin^{-1}u+(1-u^2)}^{1/2}$ substiute back,
$\displaystyle 2x\sin^{-1}(2x)+\sqrt{1-4x^2}+C$
Remember to mutiply by 1/2 as in the beginning of problem,
$\displaystyle x\sin^{-1}2x+\frac{1}{2}\sqrt{1-4x^2}+C$
Q.E.D.
• Feb 26th 2006, 10:31 AM
alysha27
Thanks, I hate little mistakes like that..