How can I find $\int(\sin ^4 x ) dx$? 
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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?
$$\int(\sin ^4 x ) dx$$
Thanks in advance.
 A: 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.
A: Sometimes handling trigonometric functions is easier in complex exponential form:
$$\sin ^4(x)$$
$$\frac{1}{16} \left(e^{-i x}-e^{i x}\right)^4$$
$$-\frac{1}{4} e^{-2 i x}-\frac{1}{4} e^{2 i x}+\frac{1}{16} e^{-4 i x}+\frac{1}{16} e^{4
   i x}+\frac{3}{8}$$
$$-\frac{1}{2} \cos (2 x)+\frac{1}{8} \cos (4 x)+\frac{3}{8}$$
A: 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 $$\sin^{2}(x) = \frac{1}{2}-\frac{1}{2} \cos(2x)$$
A: 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.)
$\cos 4x=\cos^4 x- 6\cos^2x\sin^2x + \sin^4 x = (1-\sin^2x)^2-6(1-\sin^2x)\sin^2x+\sin^4x=\\=1-2\sin^2x+\sin^4x-6\sin^2x+6\sin^4x+\sin^4x-8\sin^4x-8\sin^2x+1$
$\cos 2x = \cos^2x - \sin^2x=1-2\sin^2x$
$\sin^4x=\sin^2x+\frac{\cos4x-1}8=\frac{1-\cos2x}2+\frac{\cos4x-1}8$
