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

The integral of sin^2(θ)cos(4θ) is -1/8 (1/2 sin(2θ)-1/6 sin(6θ))+1/4 sin^2(θ)sin(4θ)+C

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integral of sin^2(θ)cos(4θ)

\int sin^{2} (θ) cos (4 θ) d θ

- \frac{1}{8} (\frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ)) + \frac{1}{4} sin^{2} (θ) sin (4 θ) + C

Solution steps

\int sin^{2} (θ) cos (4 θ) d θ

Rewrite using trig identities

= \int (1 - cos^{2} (θ)) cos (4 θ) d θ

Apply Integration By Parts

: - \int \frac{1}{4} sin (2 θ) sin (4 θ) d θ + \frac{1}{4} sin^{2} (θ) sin (4 θ)

= - \int \frac{1}{4} sin (2 θ) sin (4 θ) d θ + \frac{1}{4} sin^{2} (θ) sin (4 θ)

\int \frac{1}{4} sin (2 θ) sin (4 θ) d θ = \frac{1}{8} (\frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ))

= - \frac{1}{8} (\frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ)) + \frac{1}{4} sin^{2} (θ) sin (4 θ)

Add a constant to the solution

= - \frac{1}{8} (\frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ)) + \frac{1}{4} sin^{2} (θ) sin (4 θ) + C

n = 1 \sum \infty \frac{3 n !}{n ^{n}}

\int_{0}^{\frac{π}{2}} sin (2 x) d x

3 t^{4} - 2 t^{2} + 1

cos (2 t)

y = 16 - x^{2}

What is the integral of sin^2(θ)cos(4θ) ?
The integral of sin^2(θ)cos(4θ) is -1/8 (1/2 sin(2θ)-1/6 sin(6θ))+1/4 sin^2(θ)sin(4θ)+C