# Solving $\int \sin^6(x)\cos^2(x)dx$

$$\int \sin^6(x)\cos^2(x)dx$$ $$t=\sin x$$ or $$\cos x$$ doesn't work obviously

In general, how do I approach the integrals of the form - $$\int \sin^m(x)\cos^n(x)dx ; x,y \in 2n, n\in \mathcal{I^+}$$

I'm ruling out the possibility of taking $$\cos^2x=1-\sin^2x$$ and applying sine reduction formula, since its just tedious for higher $$m$$(s)

Edit: I'm not "accepting" any answer, since every answer is equally good and I'm trying to get as many approaches/answers as possible :-)

• Sometimes tedium is the only way. Perhaps it is worth a try to see what turns up... – abiessu Sep 9 '20 at 12:53
• @PeterForeman Im sorry, Fourier series is way beyond the scope of my knowledge – DatBoi Sep 9 '20 at 12:54
• @abiessu thats double the work. First we have to expand and multiply a lot, and then apply reduction formula for so many terms. – DatBoi Sep 9 '20 at 12:56

## 6 Answers

Bioche's rules recommend $$u=\tan x$$, giving$$\int\sin^6x\cos^2xdx=\int\sin^6x\cos^4xdu=\int\tan^6x\cos^{10}xdu=\int\frac{u^6du}{(1+u^2)^5}.$$Now it's an exercise in partial fractions. No matter how you do it, the result is messy.

• Thank you for the answer. It does get really messy! – DatBoi Sep 9 '20 at 13:32
• +1 for Bioche's rules, also if I prefer the answer by @zwim – enzotib Sep 9 '20 at 14:08
• @enzoib Come to think of it, so do I, even though the answer by sirous has the advantage of not requiring the reader to know about complex numbers. – J.G. Sep 9 '20 at 14:10
• In the case of a single $\sin^n x$ or $\cos^n x,$ with even $n,$ me too try to use duplication formulae, but for a product of even powers it becomes quickly a mess – enzotib Sep 9 '20 at 14:15

One possibility is to linearize this expression using $$\begin{cases}\cos(x)^2=\left(\frac{e^{ix}+e^{-ix}}{2i}\right)^2\\\sin(x)^6=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^6\end{cases}$$

You will get a sum of $$e^{\pm nx}$$ terms than you can then recombine to get $$\cos(nx)$$ or $$\sin(nx)$$ terms.

$$-\frac 1{256}e^{8ix}+\frac 1{64}e^{6ix}-\frac 1{64}e^{4ix}-\frac 1{64}e^{2ix}+\frac 5{128}-\frac 1{64}e^{-2ix}-\frac 1{64}e^{-4ix}+\frac 1{64}e^{-6ix}-\frac 1{256}e^{-8ix}$$

$$= -\frac 1{128}\cos(8x)+\frac 1{32}\cos(6x)-\frac 1{32}\cos(4x)-\frac 1{32}\cos(2x)+\frac 5{128}$$

Now this is straightforward to integrate.

• That's elegant! Thank you! – DatBoi Sep 9 '20 at 13:35

The standard methods for monomials is $$\sin$$ and $$\cos$$ depends on the values of the exponents:

• if the exponent of $$\sin$$ is odd, set $$t=\cos x,\;\mathrm dt=-\sin x\,\mathrm dx$$. If the exponent of $$\cos$$ is odd, set $$t=\sin x,\;\mathrm dt=\sin x\,\mathrm dx$$.
• If both exponents are even, it is simpler to linearise the monomial, either with pure trigonometry or with complex numbers. I'll show how it goes with the latter method:

To simplify the computations, we'll denote $$u=\mathrm e^{ix},\;\bar u=\mathrm e^{-ix}$$, so that $$\sin x=\frac{u-\bar u}{2i},\quad\cos x=\frac{u+\bar u}{2}.$$ We'll also apply the following rules: $$u\,\bar u=1,\quad u^k+\bar u^k=2\cos kx,,\quad u^k-\bar u^k=2i\sin kx.$$ Now the given monomial becomes \begin{align} \sin^6x\cos^2x& =\frac{(u-\bar u)^6}{(2i)^6}\,\frac{(u+\bar u)^2}{2^2}=-\frac{(u-\bar u)^4(u^2-\bar u)^2}{256}\\ &= -\frac{(u^4-4u^2+6-4\bar u^2+\bar u^4)(u^4-2+\bar u^4)}{256}\\ &=-\frac1{256}\Bigl[(u^8-4u^6+6u^4-4u^2+1)-(2u^4-8u^2+12-8\bar u^2+2\bar u^4)\\ &\hspace{6em}+(1-4\bar u^2+6\bar u^4-4\bar u^6 +\bar u^8\Bigr] \\ &=-\frac1{128}\bigl[\cos 8x-4\cos 6x+4\cos 4x +4\cos 2x -5\bigr]. \end{align}

Hint:

If you are patient enough you can use following identities:

$$\cos^2 (x)=\frac{1}2(1+\cos (2x))$$

$$\sin^6(x)=\frac{1}{16}[9(1-\cos 2x)/2+(1-\cos 6x)/2-3(\cos 2x-\cos 4x)]$$

Now multiply and reduce the product to algebraic sum of $$\sin kx$$ or $$\cos tx$$ and integrate.

• This is very similar to the method proposed by @zwim but without complex numbers! – DatBoi Sep 9 '20 at 14:13
• @DatBoi, just wanted to show the method for this particular case.. of course method proposed by zwin is a general solution and is great for linearization. – sirous Sep 9 '20 at 15:50
• Of course ;---) – DatBoi Sep 9 '20 at 15:56

There are a recursive formulas:

$$\int\sin^nax\cos^max\space dx =-\frac{\sin^{n-1}ax \cos^{m+1}ax}{a(n+m)}+\frac{n-1}{n+m}\int\sin^{n-2}ax \cos^m ax \space dx$$

(lowering exponent $$n$$; $$m$$ and $$n > 0$$),

$$=\frac{\sin^{n+1}ax \cos^{m-1}ax}{a(n+m)}+\frac{m-1}{n+m}\int\sin^{n}ax \cos^{m-2} ax \space dx$$

(lowering exponent $$m$$; $$m$$ and $$n>0$$).

• I assume that "lowering exponent" means the smaller of the two exponents? – DatBoi Sep 9 '20 at 12:58
• It is up to you, but sure we will use the one which will take a less time to get the solution... – Anton Vrdoljak Sep 9 '20 at 13:01
• I understand.Thank you very much for your help! – DatBoi Sep 9 '20 at 13:04
• @AntonVrdolijak Can you give me some hints on how to manipulate the integral part of integration by parts to get the desired result? – DatBoi Sep 9 '20 at 13:09
• To @DatBoi: I borrowed above formulas from Handbook of Mathematics (authors: I.N. Bronshtein · K.A. Semendyayev · G.Musiol · H.Muehlig)... – Anton Vrdoljak Sep 9 '20 at 13:35

Apply the recursion

$$I_n=\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \sin^{n - 1}{x} + \frac{n - 1}{n} I_{n-2}$$

to integrate \begin{align} &\int \sin^6x\cos^2xdx = I_6-I_8=-\frac18I_6+\frac18\cos x\sin^7x\\ =& -\frac5{48}I_4 +\frac18\cos x\left( \frac1{6}\sin^5x+\sin^7x\right)\\ =&-\frac{15}{256}I_2 +\frac18\cos x\left(\frac5{32}\sin^3x+\frac1{6}\sin^5x+\sin^7x \right)\\ =&-\frac{15}{512}x +\frac18\cos x\left(\frac{15}{64}\sin x+ \frac5{32}\sin^3x+\frac1{6}\sin^5x+\sin^7x \right)\\ \end{align} Integrating $\int \sin^n{x} \ dx$