Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Evaluate $$\int_0^1\int_0^{\cos^{-1}y}(\sin x)\sqrt{1+\sin^2x}\,dxdy.$$

Can anyone hint me how to start solving this? Or solve the whole thing if you're generous enough. :D

share|improve this question

1 Answer 1

Since we know that the integrand is independent of $y$, we can perform integration w.r.t. $y$ first. Hence the integration becomes $\int_0^{\pi/2}\int_0^{\cos{x}}(\sin x)\sqrt{1+\sin^2x}\,dydx$ then the integration becomes a lot easier.

share|improve this answer
To OP: Drawing a picture can be helpful when trying to figure out if changing the order of integration would be helpful. After the order change, this becomes a very simple $u$ substitution. –  apnorton Dec 11 '12 at 13:36
If we change the integration from $dxdy$ to $dydx$, that would mean that the bounds will change, right? How would I solve for the new boundaries? –  Rose Dec 11 '12 at 13:42
@Cameron How did you solve for the new bounds? –  Rose Dec 11 '12 at 15:04
@Rose: If you draw a picture of the region, the boundary curves are $x=0$, $y=0$, and $x=\cos^{-1}y$ (or equivalently, $y=\cos x$). If we go with $dxdy$, then we go from $x=0$ on the left to $x=\cos^{-1}y$ on the right, and then go from our lower bound $y=0$ to our upper bound, which is determined in this case by the point of intersection of $x=0$ and $x=\cos^{-1}y$, namely $y=\cos 0=1$. (cont'd) –  Cameron Buie Dec 11 '12 at 15:44
Write out the original region of integration: $0 \leq x \leq \arccos y$, $0 \leq y \leq 1$. Now we want to write this a different way. If $0 \leq y \leq 1$, then $\arccos y$ ranges from $\arccos 0 \pi/2$ to $\arccos 1 =0$. So the largest $x$ can be is $\pi/2$. So the bounds on $x$ are $0 \leq x \leq \pi/2$. Now since $y$ ranges in $[0,1]$ and this happens to be the range of $\cos x$ on $[0,\pi/2]$, we have $0 \leq y \leq \cos x$. –  neuguy Dec 11 '12 at 15:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.