What is the integral of sin^5(θ) ?

The integral of sin^5(θ) is -cos(θ)+(2cos^3(θ))/3-(cos^5(θ))/5+C

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integral of sin^5(θ)

\int sin^{5} (θ) d θ

- cos (θ) + \frac{2 cos ^{3} ( θ )}{3} - \frac{cos ^{5} ( θ )}{5} + C

Solution steps

\int sin^{5} (θ) d θ

Rewrite using trig identities

= \int (1 - cos^{2} (θ))^{2} sin (θ) d θ

Apply u-substitution

= \int - 1 + 2 u^{2} - u^{4} d u

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 u + \int 2 u^{2} d u - \int u^{4} d u

\int 1 d u = u

\int 2 u^{2} d u = \frac{2 u ^{3}}{3}

\int u^{4} d u = \frac{u ^{5}}{5}

= - u + \frac{2 u ^{3}}{3} - \frac{u ^{5}}{5}

Substitute back

u = cos (θ)

= - cos (θ) + \frac{2 cos ^{3} ( θ )}{3} - \frac{cos ^{5} ( θ )}{5}

Add a constant to the solution

= - cos (θ) + \frac{2 cos ^{3} ( θ )}{3} - \frac{cos ^{5} ( θ )}{5} + C

What is the integral of sin^5(θ) ?
The integral of sin^5(θ) is -cos(θ)+(2cos^3(θ))/3-(cos^5(θ))/5+C