What is the integral of tan(θ)4sin(2θ) ?

The integral of tan(θ)4sin(2θ) is 4(θ-1/2 sin(2θ))+C

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integral of tan(θ)4sin(2θ)

\int tan (θ) 4 sin (2 θ) d θ

4 (θ - \frac{1}{2} sin (2 θ)) + C

Solution steps

\int tan (θ) \cdot 4 sin (2 θ) d θ

Take the constant out:

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

= 4 \cdot \int tan (θ) sin (2 θ) d θ

Rewrite using trig identities

= 4 \cdot \int 2 cos (θ) sin (θ) tan (θ) d θ

Take the constant out:

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

= 4 \cdot 2 \cdot \int cos (θ) sin (θ) tan (θ) d θ

Rewrite using trig identities

= 4 \cdot 2 \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

= 4 \cdot 2 \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

= 4 \cdot 2 \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)

= 4 \cdot 2 \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ))

Simplify

4 \cdot 2 \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

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What is the integral of tan(θ)4sin(2θ) ?
The integral of tan(θ)4sin(2θ) is 4(θ-1/2 sin(2θ))+C