What is the integral of 1+2sin(θ)+sin^2(θ) ?

The integral of 1+2sin(θ)+sin^2(θ) is 3/2 θ-1/4 sin(2θ)-2cos(θ)+C

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integral of 1+2sin(θ)+sin^2(θ)

\int 1 + 2 sin (θ) + sin^{2} (θ) d θ

\frac{3}{2} θ - \frac{1}{4} sin (2 θ) - 2 cos (θ) + C

Solution steps

\int 1 + 2 sin (θ) + sin^{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

= \int 1 d θ + \int 2 sin (θ) d θ + \int sin^{2} (θ) d θ

\int 1 d θ = θ

\int 2 sin (θ) d θ = - 2 cos (θ)

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

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

Simplify

θ - 2 cos (θ) + \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) : \frac{3}{2} θ - \frac{1}{4} sin (2 θ) - 2 cos (θ)

= \frac{3}{2} θ - \frac{1}{4} sin (2 θ) - 2 cos (θ)

Add a constant to the solution

= \frac{3}{2} θ - \frac{1}{4} sin (2 θ) - 2 cos (θ) + C

sin (3 x)

\int 4 e^{c o s (x)} sin (x) d x

\int \frac{2 sin ( x )}{cos ( x )} d x

f (x) = tan (8 x)

\int \frac{1}{( x ^{2} + 4 ) ( x + 2 )} d x

What is the integral of 1+2sin(θ)+sin^2(θ) ?
The integral of 1+2sin(θ)+sin^2(θ) is 3/2 θ-1/4 sin(2θ)-2cos(θ)+C