I'm stuck at calculating this integral
$\int \cos^3 y \text{ } \sin^4 y \, \mathrm{d} y$
I tried a lot of things
- $u = \cos^3(y)$, $\mathrm{d}y = -3\sin y \cos^2 y$
- $u = \sin^2(y)$, $\mathrm{d}y = 2 \sin \cos y$
- $u = \sin(y)$, $\mathrm{d}y = \sin(2y)$
- played with $\operatorname{cosec}$ and $\operatorname{sec}$
None of this worked. Do you have a hint on how to start?
Thanks to Gerry Myerson, I have a hint on how to start the problem. I am still stuck though. Sorry, I'm starting with integrals!
Here is what I've done :
$I = \int \! \cos^3(y) \sin^4(y) \, \mathrm{d} y$
$I = \int \! \cos^2(y) \cos(y)\sin^4(y) \, \mathrm{d} y$
$I = \int \! (1-\sin^2(y))\sin^4(y)\cos(y) \, \mathrm{d} y$
$u = \sin y$, so $ dy = du/ \cos y$
$I = \int \! (1-u^2)u^2 \, \mathrm{d}u$
$I = \int \! u^2-u^6 \, \mathrm{d}u$
$I = u^5/5 - u^7 / 7$
$I = (\sin y)^5/5 - (\sin y)^7 / 7$
This is definitely not the good answer...