What is the integral of 8cos^2(θ) ?

The integral of 8cos^2(θ) is 4(θ+1/2 sin(2θ))+C

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integral of 8cos^2(θ)

\int 8 cos^{2} (θ) d θ

4 (θ + \frac{1}{2} sin (2 θ)) + C

Solution steps

\int 8 cos^{2} (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^{2} (θ) d θ

Rewrite using trig identities

= 8 \cdot \int \frac{1 + cos ( 2 θ )}{2} d θ

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int 1 + cos (2 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 \cdot \frac{1}{2} (\int 1 d θ + \int cos (2 θ) d θ)

\int 1 d θ = θ

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

= 8 \cdot \frac{1}{2} (θ + \frac{1}{2} sin (2 θ))

Simplify

8 \cdot \frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) : 4 (θ + \frac{1}{2} sin (2 θ))

= 4 (θ + \frac{1}{2} sin (2 θ))

Add a constant to the solution

= 4 (θ + \frac{1}{2} sin (2 θ)) + C

\int \frac{2 x + 1}{x ( x + 1 )} d x

\int sec (w) d w

\int \frac{1}{( 1 + 3 x )} d x

\int 4 x I n (8 x + 6)

\int \frac{x ^{2} + 2 x - 1}{3 - x} d x

What is the integral of 8cos^2(θ) ?
The integral of 8cos^2(θ) is 4(θ+1/2 sin(2θ))+C