What is the integral of 2cos(x)-sin^3(x)cos^3(x) ?

The integral of 2cos(x)-sin^3(x)cos^3(x) is 2sin(x)+1/4 cos^4(x)-1/6 cos^6(x)+C

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integral of 2cos(x)-sin^3(x)cos^3(x)

integral

2 cos (x) - sin^{3} (x) cos^{3} (x)

2 sin (x) + \frac{1}{4} cos^{4} (x) - \frac{1}{6} cos^{6} (x) + C

Solution steps

\int 2 cos (x) - sin^{3} (x) cos^{3} (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 2 cos (x) d x - \int sin^{3} (x) cos^{3} (x) d x

\int 2 cos (x) d x = 2 sin (x)

\int sin^{3} (x) cos^{3} (x) d x = - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}

= 2 sin (x) - (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6})

Simplify

2 sin (x) - (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}) : 2 sin (x) + \frac{1}{4} cos^{4} (x) - \frac{1}{6} cos^{6} (x)

= 2 sin (x) + \frac{1}{4} cos^{4} (x) - \frac{1}{6} cos^{6} (x)

Add a constant to the solution

= 2 sin (x) + \frac{1}{4} cos^{4} (x) - \frac{1}{6} cos^{6} (x) + C

\frac{1}{( 3 x - 1 )}

x^{2} (x^{3} + 2)^{4}

\frac{( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )}

\frac{- cos ^{2} ( x )}{2}

1. 3^{x}

What is the integral of 2cos(x)-sin^3(x)cos^3(x) ?
The integral of 2cos(x)-sin^3(x)cos^3(x) is 2sin(x)+1/4 cos^4(x)-1/6 cos^6(x)+C