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Possible Duplicate:
Evaluating $\\int P(\\sin x, \\cos x) \\text{d}x$

Hi,

My question is: How can I solve the following integral question? $\int(\sin ^4 x ) dx$

Thanks in advance.

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    Whoever answered this question might want to give a general answer to this "abstract duplicate": http://math.stackexchange.com/questions/29980/evaluating-int-p-sin-x-cos-x-textdx2011-03-30

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There are standard techniques for solving integrals that consist of powers of sines and cosines.

One is to use reduction formulas when you simply have a power of sines or a power of cosines. These can be obtained by performing integration by parts, followed by using a trigonometric identity. This is done in pretty much every single calculus textbook I have ever encountered.

Another is to use trigonometric power reduction formulas to change the fourth power of the sine into a sum of multiples of simple cosines, which can then be solved with an easy substitution.

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Sometimes handling trigonometric functions is easier in complex exponential form:

$\sin ^4(x)$ $\frac{1}{16} \left(e^{-i x}-e^{i x}\right)^4$ $-\frac{1}{4} e^{-2 i x}-\frac{1}{4} e^{2 i x}+\frac{1}{16} e^{-4 i x}+\frac{1}{16} e^{4 i x}+\frac{3}{8}$ $-\frac{1}{2} \cos (2 x)+\frac{1}{8} \cos (4 x)+\frac{3}{8}$

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    Note that I did not integrate the expression. I just rewrote it to make integration easier.2013-02-20
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Note that $\int \sin^{n} \ dx = -\frac{\cos(x) \sin^{n-1}(x)}{n}+ \frac{n-1}{n} \int \sin^{n-2}(x) \ dx$ and $\sin^{2}(x) = \frac{1}{2}-\frac{1}{2} \cos(2x)$

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    Note that the reductio$n$ formulas are typically not something one would *note* unless you've already seen them, in which case you probably have seen and know how to approach problems like the one posted. The double-angle identities, on the other hand, are certainly helpful for this particular integral.2011-03-30
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This can be viewed as almost the same thing as exponential form, only in this case I assume that you remember multiple angles formula. (I.e. you do not need to use complex numbers.) $\cos 4x=\cos^4 x- 6\cos^2x\sin^2x + \sin^4 x = (1-\sin^2x)^2-6(1-\sin^2x)\sin^2x+\sin^4x=\\=1-2\sin^2x+\sin^4x-6\sin^2x+6\sin^4x+\sin^4x-8\sin^4x-8\sin^2x+1$

$\cos 2x = \cos^2x - \sin^2x=1-2\sin^2x$

$\sin^4x=\sin^2x+\frac{\cos4x-1}8=\frac{1-\cos2x}2+\frac{\cos4x-1}8$

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    Answer on a PSQ2018-12-08