Inverse Sine Integral

**alysha27** · February 26th 2006, 10:06 AM

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

**ThePerfectHacker** · February 26th 2006, 10:29 AM

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,
$\int\sin^{-1}(2x)dx$ use the substitution $u=2x$ then by the substitution rule we have, $\frac{1}{2}\int\sin^{-1}udu$
Now consider this integral as,
$\int(1)\sin^{-1}udu$ . Use integration by parts to get,
$u\sin^{-1}u-\int\frac{u}{\sqrt{1-u^2}}du$ .
Notice this new integral is solvable by substitution, call $v=1-u^2$ , thus,
$u\sin^{-1}u+\frac{1}{2}\int v^{-1/2}dv$ .
Thus, by the exponent rule,
$u\sin^{-1}u+v^{1/2}$ substitute back,
$u\sin^{-1}u+(1-u^2)}^{1/2}$ substiute back,
$2x\sin^{-1}(2x)+\sqrt{1-4x^2}+C$
Remember to mutiply by 1/2 as in the beginning of problem,
$x\sin^{-1}2x+\frac{1}{2}\sqrt{1-4x^2}+C$
Q.E.D.

**alysha27** · February 26th 2006, 10:31 AM

Thanks, I hate little mistakes like that..

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