What is the integral of cos(x)(6sin^2(x)-1) ?

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

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integral of cos(x)(6sin^2(x)-1)

\int cos (x) (6 sin^{2} (x) - 1) d x

- sin (x) cos (2 x) + C

Solution steps

\int cos (x) (6 sin^{2} (x) - 1) d x

Expand

cos (x) (6 sin^{2} (x) - 1) : 6 sin^{2} (x) cos (x) - cos (x)

= \int 6 sin^{2} (x) cos (x) - cos (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 6 sin^{2} (x) cos (x) d x - \int cos (x) d x

\int 6 sin^{2} (x) cos (x) d x = 2 sin^{3} (x)

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

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

Simplify

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

= - sin (x) cos (2 x)

Add a constant to the solution

= - sin (x) cos (2 x) + C

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