What is the integral of 8cos^3(θ) ?

The integral of 8cos^3(θ) is 8(sin(θ)-(sin^3(θ))/3)+C

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integral of 8cos^3(θ)

\int 8 cos^{3} (θ) d θ

8 (sin (θ) - \frac{sin ^{3} ( θ )}{3}) + C

Solution steps

\int 8 cos^{3} (θ) d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int cos^{3} (θ) d θ

Rewrite using trig identities

= 8 \cdot \int (1 - sin^{2} (θ)) cos (θ) d θ

Apply u-substitution

: \int 1 - u^{2} d u

= 8 \cdot \int 1 - u^{2} 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

= 8 (\int 1 d u - \int u^{2} d u)

\int 1 d u = u

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

= 8 (u - \frac{u ^{3}}{3})

Substitute back

u = sin (θ)

= 8 (sin (θ) - \frac{sin ^{3} ( θ )}{3})

Add a constant to the solution

= 8 (sin (θ) - \frac{sin ^{3} ( θ )}{3}) + C

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What is the integral of 8cos^3(θ) ?
The integral of 8cos^3(θ) is 8(sin(θ)-(sin^3(θ))/3)+C