What is the integral of (4cos(θ))^2 ?

The integral of (4cos(θ))^2 is 16θcos^2(θ)+16(-1/2 θcos(2θ)+1/4 sin(2θ))+C

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integral of (4cos(θ))^2

\int (4 cos (θ))^{2} d θ

16 θ cos^{2} (θ) + 16 (- \frac{1}{2} θ cos (2 θ) + \frac{1}{4} sin (2 θ)) + C

Solution steps

\int (4 cos (θ))^{2} d θ

Apply Integration By Parts

: 16 θ cos^{2} (θ) - \int - 32 θ cos (θ) sin (θ) d θ

= 16 θ cos^{2} (θ) - \int - 32 θ cos (θ) sin (θ) d θ

\int - 32 θ cos (θ) sin (θ) d θ = - 16 (- \frac{1}{2} θ cos (2 θ) + \frac{1}{4} sin (2 θ))

= 16 θ cos^{2} (θ) - (- 16 (- \frac{1}{2} θ cos (2 θ) + \frac{1}{4} sin (2 θ)))

Simplify

= 16 θ cos^{2} (θ) + 16 (- \frac{1}{2} θ cos (2 θ) + \frac{1}{4} sin (2 θ))

Add a constant to the solution

= 16 θ cos^{2} (θ) + 16 (- \frac{1}{2} θ cos (2 θ) + \frac{1}{4} sin (2 θ)) + C

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What is the integral of (4cos(θ))^2 ?
The integral of (4cos(θ))^2 is 16θcos^2(θ)+16(-1/2 θcos(2θ)+1/4 sin(2θ))+C