# Calculate the integral of $\cos^3(y)\sin^4(y)\,\mathrm{d}y$

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...

• What do you mean by "not the good answer"? Mar 15, 2012 at 3:23
• If I go to Wolfram Alpha, I get this answer : $1/70 sin^5(y) (5 cos(2 y)+9)+constant$ Mar 15, 2012 at 3:26
• I mean that I think I made a mistake somewhere. Mar 15, 2012 at 3:43
• Well a friend calculated an integral and obtained $\sin^2(x)$ but WA yield $1-\cos^2(x)$, so he probably made a mistake, right? Mar 15, 2012 at 3:54
• @JustinDomingue In other words: Hint: $(5 \cos(2y)+9)=5 -10 \sin^2(y)+9$. Mar 15, 2012 at 3:56

$\cos^2y=1-\sin^2y$; $u=\sin y$.
Your answer is close to perfect. First note that you missed a integration constant $C$ in the last two lines and then check that $$\begin{eqnarray} \frac{d}{dx}\left(\frac{(\sin x)^5}{5}-\frac{(\sin x)^7}{7}+C\right)&=&(\sin x)^4\cos x - (\sin x)^6\cos x\\ &=&(\sin x)^4\cos x(1-(\sin x)^2)\\ &=&(\sin x)^4(\cos x)^3\\ \end{eqnarray}$$