Question

How to integrate ∫▒〖(sinx)〗^4 .〖(cosx)〗^4.dx?

Answer 1

The integral:

$I= \int\limits^{}_{} {Sin^4x\ Cos^4x} \, dx\\\\=\frac{1}{2^4}\int\limits^{}_{} {Sin^42x} \, dx\\\\=\frac{1}{2^4}\int\limits^{}_{} {(Sin^22x)^2} \, dx\\\\=\frac{1}{2^4}\frac{1}{2^2}\int\limits^{}_{} {(1-Cos4x)^2} \, dx\\\\=\frac{1}{2^6}\int\limits^{}_{} {(1+Cos^24x-2\ Cos4x)} \, dx\\\\=\frac{1}{2^6}\int\limits^{}_{} {[1+\frac{1}{2}(1+Cos8x)-2\ Cos4x]} \, dx\\\\=\frac{1}{2^6} \int\limits^{}_{} {\frac{3}{2}} \, dx +\frac{1}{2^7} \int\limits^{}_{} {Cos8x} \, dx -\frac{1}{2^5} \int\limits^{}_{} {Cos4x} \, dx$

$I=\frac{3}{2^7}x+\frac{1}{2^{10}}Sin8x-\frac{1}{2^7}Sin4x+K$

there could be other ways. this is one.

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Another way:

we know that  Sin 3x = 3 Sin x - 4 SIn³ x
=>    Sin x Sin 3x = 3 Sin²x - 4 Sin⁴ x
=>    1/2 (Cos 2x - Cos 4x) = 3/2 (1-Cos2x) - 4 Sin⁴ x
=>  Sin⁴ x = 1/8 [ 3 - 3 Cos 2x + Cos 4x - Cos 2x ]
=>  Sin⁴ 2x = 1/8 [ 3 - 4 Cos 4x + Cos 8x ]

$I= \int\limits^{}_{} {Sin^4x\ Cos^4x} \, dx\\\\=\frac{1}{2^4}\int\limits^{}_{} {Sin^42x} \, dx\\\\=\frac{1}{2^7}\int\limits^{}_{} {[ 3 - 4 Cos4x + Cos 8x ] } \, dx\\\\=\frac{3}{2^7}x-\frac{1}{2^7}Sin4x+\frac{1}{2^{10}}Sin8x+K$

This is simpler, if you remember formula for sin 3x or sin 4x.