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I need to find the antiderivative of $$\int\sin^6x\cos^2x \mathrm{d}x.$$ I tried symbolizing $u$ as squared $\sin$ or $\cos$ but that doesn't work. Also I tried using the identity of $1-\cos^2 x = \sin^2 x$ and again if I symbolize $t = \sin^2 x$ I'm stuck with its derivative in the $dt$.

Can I be given a hint?

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    $\begingroup$ When you say the antiderivative of the integral, do you just mean the indefinite integral? Or do you mean take the antiderivative of the indefinite integral? $\endgroup$ – Ian Coley Mar 16 '13 at 8:13
  • $\begingroup$ When I said antiderivative of the aforemetioned integral I meant the function F, which its derivative is sin^6(x)*cos^2(x). $\endgroup$ – user65985 Mar 16 '13 at 8:16
  • $\begingroup$ You should try linearize $\sin^6(x)\cos^2(x)$, that is express it as a linear combination of $\cos(kx)$ and $\sin(kx)$ ($k$ integer). $\endgroup$ – Siméon Mar 16 '13 at 8:18
  • $\begingroup$ that's problematic. The function inside the integral is (1-cos(2x))^3*(cos(2x)+1)/16 but still I stuck with a lot of cos(kx) in various exponents smaller than 4... $\endgroup$ – user65985 Mar 16 '13 at 8:23
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    $\begingroup$ You should find: $$\sin^6 x\cos^2 x = \frac{5-4\cos(2x)-4\cos(4x) + 4\cos(6x) -\cos(8x) }{128} $$ $\endgroup$ – Siméon Mar 16 '13 at 8:26
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Hint We can use double-angle identities to reduce powers. We could use $\cos 2t=2\cos^2 t-1$ and $\cos 2t=1-2\sin^2 t$. We end up with polynomial of degree $4$ in $\cos 2x$. Repeat the idea where needed.

It is more efficient in this case to use $\sin 2t=2\sin t\cos t$, that is, first rewrite our expression as $(\sin x\cos x)^2\sin^4 x$. Then rewrite as $\frac{1}{16}(\sin^2 2x)(1-\cos 2x)^2$. Expand the square. Not quite finished. But we end up having to integrate $\sin^2 2x$ (standard), $\sin^2 2x\cos 2x$ (simple substitution), and $\sin^2 2x\cos^2 2x$, a close relative of $\sin^2 4x$.

Remark: In this problem, like in a number of trigonometric integrations, it is possible to end up with dramatically different-looking answers. They all differ by constants, so we are saved by the $+C$.

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  • $\begingroup$ how do you subtitute sin^2(2X)cos(2x)? $\endgroup$ – user65985 Mar 16 '13 at 8:51
  • $\begingroup$ Let $u=\sin 2x$. Then $du=2\cos 2x\,dx$. So we end up with $\int\frac{1}{2}u^2\,du$. $\endgroup$ – André Nicolas Mar 16 '13 at 8:53
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Just adding a good point, however, you got the answer completely $\ddot\smile$ :

Consider $$\int\sin^m(x)\cos^n(x)dx$$ where in $m,n\in\mathbb Q$. Whenever $m+n$ is an even integer, you can use $t=\tan(x)$ or $t=\cos(x)$ as a good substitution.

And here $m+n=8$ is an even integer.

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  • $\begingroup$ And a good point it is, +1. $\endgroup$ – Julien Mar 16 '13 at 17:17
  • $\begingroup$ @julien: Thanks julien. I was sure after 9 hours, nobody would visit this answer but you did. :-) $\endgroup$ – mrs Mar 16 '13 at 17:21
  • $\begingroup$ And I visited this answer too! +1 $\endgroup$ – Namaste Mar 17 '13 at 0:47
  • $\begingroup$ You ARE always Welcome! ;-) $\endgroup$ – mrs Mar 17 '13 at 4:59
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$$\text{ As }\cos2y=2\cos^2y-1=1-2\sin^2y$$

$$\sin^6x\cos^2x=\left(\frac{1-\cos2x}2\right)^3\left(\frac{1+\cos2x}2\right)$$

$$16\sin^6x\cos^2x=(1-3\cos2x+3\cos^2x-\cos^32x)(1+\cos2x)$$

$$=\left(1-3\cos2x+3\frac{(1+\cos4x)}2-\frac{(\cos6x+3\cos2x)}4\right)(1+\cos2x)$$ (applying $\cos3y=4\cos^3y-3\cos y$)

$$64\sin^6x\cos^2x=(10-15\cos2x+6\cos4x-\cos6x)(1+\cos2x)$$

$$=10-15\cos2x+6\cos4x-\cos6x+10\cos2x-15\cos^22x+6\cos4x\cos2x-\cos6x\cos2x$$

$$=10-5\cos2x+6\cos4x-\cos6x+10\cos2x-15\frac{(1+\cos4x)}2+6\frac{(\cos2x+\cos6x)}2-\frac{(\cos4x+\cos8x)}2$$ (applying $2\cos A\cos B=\cos(A-B)+\cos(A+B)$)

$$\text{ So, }128\sin^6x\cos^2x=5-4\cos2x-4\cos4x+4\cos6x-\cos8x$$


Alternatively

as we know, $e^{ix}=\cos x+i\sin x,e^{-ix}=\cos x-i\sin x\implies \cos x=\frac{e^{ix}+e^{-ix}}2,\sin x=\frac{e^{ix}-e^{-ix}}{2i}$

So, $$\sin^6x\cos^2x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^6\left(\frac{e^{ix}+e^{-ix}}2\right)^2$$

$$=\frac{\left(e^{6ix}+e^{-6ix}-\binom61(e^{4ix}+e^{-4ix})+\binom62(e^{2ix}+e^{-2ix})-\binom63\right)}{-2^6}$$ $$\cdot\frac{\left(e^{2ix}+e^{-2ix}+2\right)}{2^2}$$

$$=\frac{e^{8ix}+e^{-8ix}-(6-2)(e^{6ix}+e^{-6ix})+(1+\binom62-2\cdot\binom61)(e^{4ix}+e^{-4ix})-(\binom63-2\cdot\binom62+\binom61)(e^{2ix}+e^{-2ix})+2\binom62-2\binom63}{-2^8}$$

$$=\frac{2\cos8x-4\cdot2\cos6x+4\cdot2\cos4x+4\cdot2\cos2x-10}{-256}\text{ as }e^{ix}+e^{-ix}=2\cos x$$

Now, simplify and use $\int\cos mxdx=\frac{\sin mx}m+C$

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$$\int \sin^6x\cos^2xdx=\int \sin^6x(1-\sin^2x)dx=\int \sin^6xdx-\int \sin^8xdx$$ $$=I_6-I_8 \text{ where }I_n=\int\sin^nxdx$$

$$\text{Now, }I_{n+2}=\int\sin^{n+2}xdx=\int\sin^{n+1}x\cdot \sin xdx$$ $$=\sin^{n+1}x\int \sin xdx-\int\left(\frac{d \sin^{n+1}x}{dx} \int \sin xdx\right)dx$$ (using Integration by parts) $$=\sin^{n+1}x(-\cos x)-\int(n+1) \sin^nx\cos x(-\cos x)dx$$ $$=-\sin^{n+1}x\cos x+(n+1)\int \sin^nx(1-\sin^2x)dx$$ $$=-\sin^{n+1}x\cos x+(n+1)(I_n-I_{n+2})$$ $$\implies I_{n+2}=-\frac{\sin^{n+1}x\cos x}{(n+2)}+\frac{n+1}{n+2}I_n$$

Now, $I_0=\int \sin^0xdx=\int dx=x$

Put $n=0,2,4,6$ respectively to get the values of $I_2,I_4,I_6,I_8$

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  • $\begingroup$ This reduction formula can be used if at least one the positive powers $\cos x,\sin x$ is even. $\endgroup$ – lab bhattacharjee Mar 17 '13 at 15:10
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If you only want a hint that will be simple:

Whenever we have even powers of sin and cos multiplied then we must convert the integral into higher angles.

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