What is the integral of 8cos^3(5x) ?

The integral of 8cos^3(5x) is 8/5 (sin(5x)-1/3 sin^3(5x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8cos^3(5x)

\int 8 cos^{3} (5 x) d x

\frac{8}{5} (sin (5 x) - \frac{1}{3} sin^{3} (5 x)) + C

Solution steps

\int 8 cos^{3} (5 x) d x

Take the constant out:

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

= 8 \cdot \int cos^{3} (5 x) d x

Apply u-substitution

: \int cos^{3} (u) \frac{1}{5} d u

= 8 \cdot \int cos^{3} (u) \frac{1}{5} d u

Take the constant out:

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

= 8 \cdot \frac{1}{5} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 8 \cdot \frac{1}{5} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int 1 - v^{2} d v

= 8 \cdot \frac{1}{5} \cdot \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 \cdot \frac{1}{5} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= 8 \cdot \frac{1}{5} (v - \frac{v ^{3}}{3})

Substitute back

= 8 \cdot \frac{1}{5} (sin (5 x) - \frac{sin ^{3} ( 5 x )}{3})

Simplify

8 \cdot \frac{1}{5} (sin (5 x) - \frac{sin ^{3} ( 5 x )}{3}) : \frac{8}{5} (sin (5 x) - \frac{1}{3} sin^{3} (5 x))

= \frac{8}{5} (sin (5 x) - \frac{1}{3} sin^{3} (5 x))

Add a constant to the solution

= \frac{8}{5} (sin (5 x) - \frac{1}{3} sin^{3} (5 x)) + C

\int cot (2 x) (cot (2 x) + 3 csc (2 x)) d x

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

\int \frac{e ^{- 2 t}}{( 1 + e ^{- t} )} d t

\int \frac{2 ^{l n (x^{3})}}{16 x} d x

\int (5 x + 7)^{4} d x

What is the integral of 8cos^3(5x) ?
The integral of 8cos^3(5x) is 8/5 (sin(5x)-1/3 sin^3(5x))+C