1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Using sort of "brute force" you can break the exponential sinusoids down into a linear combination of sinusoids. $\endgroup$
    – Jared
    May 6 '14 at 22:39
4
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Ok I don't see where this is going. Do I substitute now? $\endgroup$
    – Brian
    May 6 '14 at 22:44
  • $\begingroup$ See the answer below. Indeed, I was hinting to a substitution. $\endgroup$
    – DanielC
    May 6 '14 at 22:47
2
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ Thanks but I don't understand how sin^(3)x became sin^(2)xd(cosx) along with the negative? $\endgroup$
    – Brian
    May 6 '14 at 22:50
  • $\begingroup$ @BrianP.: Notice that $\frac{d}{dx}(\cos x)=-\sin x$ so $\sin xdx=-d(\cos x)$ $\endgroup$
    – Teddy Frei
    May 6 '14 at 22:53
  • $\begingroup$ 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? $\endgroup$
    – Brian
    May 6 '14 at 23:01
  • $\begingroup$ @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 $u=\cos x$. $\endgroup$
    – Teddy Frei
    May 6 '14 at 23:07
  • $\begingroup$ OHHHH, now it all makes sense to me. Thank you very much! $\endgroup$
    – Brian
    May 6 '14 at 23:09
1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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