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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( x - 4 ) ^{3}}{x - 3} d x

in t e g r a l \frac{1}{x - 1}

\int (x^{\frac{2}{5}} + \frac{4}{x} - 16) d x

\int (x^{2} - 4)^{- \frac{3}{2}} d x

\int \frac{( 4 x + 7 )}{x ^{2} - 5 x + 8} d x

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