Find the indefinite integral of...?

I have spent well over an hour trying to solve this equation: $$\int\cos^4{x}\sin^3{x}\, dx$$

I have tried substituting u as $\cos{x}$, $\sin{x}$, $\cos^4{x}$, and $sin^3{x}$ to no avail. How can I solve this?

• Using sort of "brute force" you can break the exponential sinusoids down into a linear combination of sinusoids. May 6 '14 at 22:39

Write $\sin^3 x = \sin x (1-\cos^2 x)$

• Ok I don't see where this is going. Do I substitute now? May 6 '14 at 22:44
• See the answer below. Indeed, I was hinting to a substitution. May 6 '14 at 22:47

$$\int \cos^4 x\sin^3 x\mathrm{d}x=-\int\cos^4x\sin^2x\mathrm{d}(\cos x)=-\int\cos^4x(1-\cos^2x)\mathrm{d}(\cos x)\\ =-\int(\cos^4x-\cos^6x)\mathrm{d}(\cos x)=-\int\cos^4x\mathrm{d}(\cos x)+\int\cos^6x\mathrm{d}(\cos x)\\ =-\frac{1}{5}\cos^5x+\frac{1}{7}\cos^7x+C,\ C\ \text{is an arbitrary constant}$$

• Thanks but I don't understand how sin^(3)x became sin^(2)xd(cosx) along with the negative? May 6 '14 at 22:50
• @BrianP.: Notice that $\frac{d}{dx}(\cos x)=-\sin x$ so $\sin xdx=-d(\cos x)$ May 6 '14 at 22:53
• Ahhh that makes sense. One last question, I don't understand how -$\int\cos^4{x}du$ becomes -$\frac{1}{5}\cos^5{x}$ where u = cosx? shouldn't it become some value sinx? May 6 '14 at 23:01
• @BrianP.:In fact, just as you say, $u=\cos x$, so the integral becomes $-\int u^4\mathrm{d}u+\int u^6\mathrm{d}u=-\frac{1}{5}u^5+\frac{1}{7}u^7+C$, then substitute back as $u=\cos x$. May 6 '14 at 23:07
• OHHHH, now it all makes sense to me. Thank you very much! May 6 '14 at 23:09

Through "brute force", you can just write out the values...there are several identities you need to get there:

$$\sin(a)\sin(b) = \frac{\cos(a - b)-\cos(a + b)}{2} = \frac{\cos(b - a)-\cos(a + b)}{2} \\ \cos(a)\cos(b) = \frac{\cos(a - b)+\cos(a + b)}{2} = \frac{\cos(b - a) + \cos(a + b)}{2} \\ \sin(a)\cos(b) = \frac{\sin(a + b) + \sin(a - b)}{2}$$

You can verify all of those properties from the angle addition properties:

\begin{align} \cos(a + b) =&& \cos(a)\cos(b) - \sin(a)\sin(b) \\ \cos(a - b) =& \cos(a)\cos(-b) - \sin(a)\sin(-b)= &\cos(a)\cos(b) + \sin(a)\sin(b)\\ \sin(a +b) =&& \sin(a)\cos(b) + \sin(b)\cos(a) \\ \sin(a - b) =& \sin(a)\cos(-b) + \sin(-b)\cos(a) =&\sin(a)\cos(b) - \sin(b)\cos(a) \end{align}

This gives:

\begin{align} \cos^4(x)sin^3(x) = &\cos(x)\left(\sin(x)\cos(x)\right)^3 \\ =& \cos(x)\frac{\sin^3(2x)}{2^3} = \cos(x)\sin(2x)\sin^2(2x) \\ =& \frac{1}{8}\cos(x)\sin(2x)\frac{1 - \cos(4x)}{2} \\ \cos(x)\sin(2x) =& \frac{\sin(3x) + \sin(x)}{2} \\ \cos^4(x)\sin^3(x) = & \left.\left.\frac{1}{32}\right(\sin(3x) - \sin(3x)\cos(4x) + \sin(x) - \sin(x)\cos(4x)\right) \\ =& \left.\left.\frac{1}{32}\right(\sin(3x) - \sin(x)\right) - \left.\left.\frac{1}{64} \right(\sin(7x) + \sin(x) + \sin(5x) + \sin(3x)\right) \\ =& \left.\left.\frac{1}{64}\right(-3\sin(x) + \sin(3x) - \sin(5x) - \sin(7x)\right) \end{align}

This then gives the integral easily as:

$$\int\cos^4(x)\sin^3(x)dx = \left.\left.\frac{1}{64}\right(3\cos(x) - \frac{1}{3}\cos(3x) + \frac{1}{5}\cos(5x) + \frac{1}{7}\cos(7x)\right) + C$$