What is the integral of (cos^4(x)-sin^4(x)) ?

The integral of (cos^4(x)-sin^4(x)) is 1/4 sin^3(x)cos(x)-3/32 sin(4x)+3/16 sin(2x)+cos^3(x)sin(x)+C

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integral of (cos^4(x)-sin^4(x))

\int (cos^{4} (x) - sin^{4} (x)) d x

\frac{1}{4} sin^{3} (x) cos (x) - \frac{3}{32} sin (4 x) + \frac{3}{16} sin (2 x) + cos^{3} (x) sin (x) + C

Solution steps

\int cos^{4} (x) - sin^{4} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int cos^{4} (x) d x - \int sin^{4} (x) d x

\int cos^{4} (x) d x = cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))

\int sin^{4} (x) d x = - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))

= cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x)) - (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))

Simplify

cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x)) - (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) : \frac{1}{4} sin^{3} (x) cos (x) - \frac{3}{32} sin (4 x) + \frac{3}{16} sin (2 x) + cos^{3} (x) sin (x)

= \frac{1}{4} sin^{3} (x) cos (x) - \frac{3}{32} sin (4 x) + \frac{3}{16} sin (2 x) + cos^{3} (x) sin (x)

Add a constant to the solution

= \frac{1}{4} sin^{3} (x) cos (x) - \frac{3}{32} sin (4 x) + \frac{3}{16} sin (2 x) + cos^{3} (x) sin (x) + C

\int 6 t + 4 d t

\int_{3}^{6} t^{2} ln (2 t) d t

\int sec^{2} (x) + 2 d x

\frac{\partial}{\partial x} (2 x^{2})

\frac{d y}{d x} = sin (2 x) - y cos (x)

What is the integral of (cos^4(x)-sin^4(x)) ?
The integral of (cos^4(x)-sin^4(x)) is 1/4 sin^3(x)cos(x)-3/32 sin(4x)+3/16 sin(2x)+cos^3(x)sin(x)+C