What is the integral of 9cos^2(θ) ?

The integral of 9cos^2(θ) is 9/2 (θ+1/2 sin(2θ))+C

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integral of 9cos^2(θ)

\int 9 cos^{2} (θ) d θ

\frac{9}{2} (θ + \frac{1}{2} sin (2 θ)) + C

Solution steps

\int 9 cos^{2} (θ) d θ

Take the constant out:

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

= 9 \cdot \int cos^{2} (θ) d θ

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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)

= 9 \cdot \frac{1}{2} (θ + \frac{1}{2} sin (2 θ))

Simplify

9 \cdot \frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) : \frac{9}{2} (θ + \frac{1}{2} sin (2 θ))

= \frac{9}{2} (θ + \frac{1}{2} sin (2 θ))

Add a constant to the solution

= \frac{9}{2} (θ + \frac{1}{2} sin (2 θ)) + C

[3 (3 x^{5} + 6)^{3}] (12 x^{3})

x = 9 y^{3}, x = 23 y^{2}

\frac{\partial}{\partial z} (e^{x y})

f (x) = (2 x + 1)^{5}

\frac{( 1 - x )}{( x + 2 )}

What is the integral of 9cos^2(θ) ?
The integral of 9cos^2(θ) is 9/2 (θ+1/2 sin(2θ))+C