∫sin²x cos²x dx
= ∫(1-cos²x)cos²x dx
=∫cos²xdx-⌠cos²xcos²xdx
=1/2⌠1+cos2x dx-1/2⌠[(1+cos2x)(1+cos2x)] Do the operations, distributions, arrange common numbers, and try to sort out the factors as a polynomiom. Then,
=1/2x+1/4sin2x-1/2x-1/2sin2x-1/4x-1/16sin4x
=-1/4x-1/16sin4x
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.5(x-sin(x)cos(x))+c
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
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Note that an angle should always be specified - for example, 1 - cos square x. Due to the Pythagorean formula, this can be simplified as sin square x. Note that sin square x is a shortcut of (sin x) squared.