# How can I find $\int(\sin ^4 x ) dx$? [duplicate]

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

Hi,

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

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## marked as duplicate by Qiaochu YuanMar 30 '11 at 20:25

Whoever answered this question might want to give a general answer to this "abstract duplicate": math.stackexchange.com/questions/29980/… – joriki Mar 30 '11 at 18:54

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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@ Chris: I checked this and I did not get the same result, though I am unsure if the two are equal in some way. I got the following: $\frac {\rm 3x}{\rm 8}-\frac {\rm 1}{\rm 4}{\rm{\sin(2x)}}+\frac {\rm 1}{\rm 32}{\rm{\sin(4x)}}$. – night owl Jul 5 '11 at 11:43
Note that I did not integrate the expression. I just rewrote it to make integration easier. – Chris Kuklewicz Feb 20 '13 at 8:22

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 reduction 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. – cch Mar 30 '11 at 18:32

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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