What is the integral of 1-sin^3(x) ?

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

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integral of 1-sin^3(x)

\int 1 - sin^{3} (x) d x

x + cos (x) - \frac{1}{3} cos^{3} (x) + C

Solution steps

\int 1 - sin^{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 1 d x - \int sin^{3} (x) d x

\int 1 d x = x

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

= x - (- cos (x) + \frac{cos ^{3} ( x )}{3})

Simplify

x - (- cos (x) + \frac{cos ^{3} ( x )}{3}) : x + cos (x) - \frac{1}{3} cos^{3} (x)

= x + cos (x) - \frac{1}{3} cos^{3} (x)

Add a constant to the solution

= x + cos (x) - \frac{1}{3} cos^{3} (x) + C

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