What is the integral of (5sin(2θ))^2 ?

The integral of (5sin(2θ))^2 is 25/2 (θ-1/4 sin(4θ))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5sin(2θ))^2

\int (5 sin (2 θ))^{2} d θ

\frac{25}{2} (θ - \frac{1}{4} sin (4 θ)) + C

Solution steps

\int (5 sin (2 θ))^{2} d θ

Rewrite using trig identities

= \int 100 cos^{2} (θ) sin^{2} (θ) d θ

Take the constant out:

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

= 100 \cdot \int cos^{2} (θ) sin^{2} (θ) d θ

Rewrite using trig identities

= 100 \cdot \int \frac{1 - cos ( 4 θ )}{8} d θ

Take the constant out:

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

= 100 \cdot \frac{1}{8} \cdot \int 1 - cos (4 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

= 100 \cdot \frac{1}{8} (\int 1 d θ - \int cos (4 θ) d θ)

\int 1 d θ = θ

\int cos (4 θ) d θ = \frac{1}{4} sin (4 θ)

= 100 \cdot \frac{1}{8} (θ - \frac{1}{4} sin (4 θ))

Simplify

100 \cdot \frac{1}{8} (θ - \frac{1}{4} sin (4 θ)) : \frac{25}{2} (θ - \frac{1}{4} sin (4 θ))

= \frac{25}{2} (θ - \frac{1}{4} sin (4 θ))

Add a constant to the solution

= \frac{25}{2} (θ - \frac{1}{4} sin (4 θ)) + C

\int \frac{1}{x - 8 x + 15} d x

\int (x^{2} + 3 x) d x

\frac{x}{x + 2}

\frac{d}{d x} (\frac{1}{( 1 + e ^{- x} )})

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

What is the integral of (5sin(2θ))^2 ?
The integral of (5sin(2θ))^2 is 25/2 (θ-1/4 sin(4θ))+C