What is the integral of 9cos^2(3θ) ?

The integral of 9cos^2(3θ) is 9/2 (θ+1/6 sin(6θ))+C

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integral of 9cos^2(3θ)

\int 9 cos^{2} (3 θ) d θ

\frac{9}{2} (θ + \frac{1}{6} sin (6 θ)) + C

Solution steps

\int 9 cos^{2} (3 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} (3 θ) d θ

Rewrite using trig identities

= 9 \cdot \int \frac{1}{2} (1 + cos (6 θ)) 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 (6 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 (6 θ) d θ)

\int 1 d θ = θ

\int cos (6 θ) d θ = \frac{1}{6} sin (6 θ)

= 9 \cdot \frac{1}{2} (θ + \frac{1}{6} sin (6 θ))

Simplify

9 \cdot \frac{1}{2} (θ + \frac{1}{6} sin (6 θ)) : \frac{9}{2} (θ + \frac{1}{6} sin (6 θ))

= \frac{9}{2} (θ + \frac{1}{6} sin (6 θ))

Add a constant to the solution

= \frac{9}{2} (θ + \frac{1}{6} sin (6 θ)) + C

lo g_{x} (2)

y = \frac{6}{x}, (16, \frac{6}{4})

y = 2 sin (x) cos (x)

\int 2 tan^{2} (x) + tan^{4} (x) d x

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

What is the integral of 9cos^2(3θ) ?
The integral of 9cos^2(3θ) is 9/2 (θ+1/6 sin(6θ))+C