What is the integral of 15sin^2(θ) ?

The integral of 15sin^2(θ) is 15/2 (θ-1/2 sin(2θ))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 15sin^2(θ)

\int 15 sin^{2} (θ) d θ

\frac{15}{2} (θ - \frac{1}{2} sin (2 θ)) + C

Solution steps

\int 15 sin^{2} (θ) d θ

Take the constant out:

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

= 15 \cdot \int sin^{2} (θ) d θ

Rewrite using trig identities

= 15 \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

= 15 \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

= 15 \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)

= 15 \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ))

Simplify

15 \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) : \frac{15}{2} (θ - \frac{1}{2} sin (2 θ))

= \frac{15}{2} (θ - \frac{1}{2} sin (2 θ))

Add a constant to the solution

= \frac{15}{2} (θ - \frac{1}{2} sin (2 θ)) + C

x \to 3 lim (\frac{3}{( x - 3 ) ^{2}})

\frac{d}{d x} (ln (\frac{x ^{2}}{5} + \frac{9}{x}))

y = 2 x^{2} - x, (1, 1)

\frac{d}{d x} (5 x + \frac{7}{x})

\int_{1}^{5} \frac{ln ( x )}{x ^{5}} d x

What is the integral of 15sin^2(θ) ?
The integral of 15sin^2(θ) is 15/2 (θ-1/2 sin(2θ))+C