Question

Evaluate the indefinite integral...

Answer 1

\[\int\limits_{}^{}\frac{ dx }{ \sqrt{1-x^2}\sin^{-1} x }\]

Answer 2

When the integrand contains \[\sqrt{a^2-b^2x^2}\]try a trig sub: let\[x=\frac{a}{b}\sin\theta\]so in this case, a=1, b=1\[x=\sin\theta, dx=\cos\theta d\theta\] also notice that when\[x=\sin\theta\Longleftrightarrow \theta=\sin^{-1}x\] then our integral\[\int\limits\frac{1}{\sqrt{1-x^2}\sin^{-1}x}dx=\int\limits\frac{1}{1-\theta\sin^2\theta}d\theta\]

Answer 3

missed a square root in the right side of the bottom equation; let me rewrite

Answer 4

actually the whole right side was messed up; should be\[\int\limits\frac{\cos\theta}{\sqrt{1-\sin^2\theta}\cdot\theta}d\theta\]

Answer 5

Hmm that's a strange approach.
Isn't this a simple `u-substitution`?\[\Large u=\sin^{-1}x\]\[\Large du=\frac{1}{\sqrt{1-x^2}}dx\]

Answer 6

yep that's way easier

Answer 7

spent too much time studying trig substitutions lately. My approach works, but yours is way faster and easier.

Answer 8

good ole trig subs XD heh

Answer 9

I'll finish mine for kicks :)

Answer 10

\[1-\sin^2\theta=\cos^2\theta\] so we have\[\int\limits\frac{\cos\theta}{\theta\sqrt{\cos^2\theta}}d\theta=\int\limits\frac{1}{\theta}d\theta=\ln \left| \theta \right|+C\] unsubstituting,\[=\ln{\left|\sin^{-1}x\right|}+C\]