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

The integral of cos(4θ)sqrt(2-sin(4θ)) is -1/6 (2-sin(4θ))^{3/2}+C

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integral of cos(4θ)sqrt(2-sin(4θ))

\int cos (4 θ) 2 - sin (4 θ) d θ

- \frac{1}{6} (2 - sin (4 θ))^{\frac{3}{2}} + C

Solution steps

\int cos (4 θ) 2 - sin (4 θ) d θ

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 - sin (4 θ)

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

Simplify

- \frac{1}{4} \cdot \frac{2}{3} (2 - sin (4 θ))^{\frac{3}{2}} : - \frac{1}{6} (2 - sin (4 θ))^{\frac{3}{2}}

= - \frac{1}{6} (2 - sin (4 θ))^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (2 - sin (4 θ))^{\frac{3}{2}} + C

k = 1 \sum \infty \frac{1}{k + 2}

1 + 3 x

e^{1 - 2 t}

\frac{d}{d x} (\frac{2 x ^{3} - 6 π ^{2}}{x ^{5}})

\int π cos (\frac{n x}{2}) d x

What is the integral of cos(4θ)sqrt(2-sin(4θ)) ?
The integral of cos(4θ)sqrt(2-sin(4θ)) is -1/6 (2-sin(4θ))^{3/2}+C