What is the integral of (sin(5x))/(e^{cos(5x)-1)} ?

The integral of (sin(5x))/(e^{cos(5x)-1)} is 1/(5e^{cos(5x)-1)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin(5x))/(e^{cos(5x)-1)}

\int \frac{sin ( 5 x )}{e ^{c o s (5 x) - 1}} d x

\frac{1}{5 e ^{c o s (5 x) - 1}} + C

Solution steps

\int \frac{sin ( 5 x )}{e ^{c o s (5 x) - 1}} d x

Apply u-substitution

: \int - \frac{1}{5 e ^{u}} d u

= \int - \frac{1}{5 e ^{u}} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int \frac{1}{e ^{u}} d u

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{5} (- \frac{1}{v})

Substitute back

= - \frac{1}{5} (- \frac{1}{e ^{c o s (5 x) - 1}})

Simplify

- \frac{1}{5} (- \frac{1}{e ^{c o s (5 x) - 1}}) : \frac{1}{5 e ^{c o s (5 x) - 1}}

= \frac{1}{5 e ^{c o s (5 x) - 1}}

Add a constant to the solution

= \frac{1}{5 e ^{c o s (5 x) - 1}} + C

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

x e^{- x + 1} 1

(81) (8 - 7)

y = 12 - 7 x - x^{2}, y = 4

cos (y x^{3}) = \frac{1}{6}

What is the integral of (sin(5x))/(e^{cos(5x)-1)} ?
The integral of (sin(5x))/(e^{cos(5x)-1)} is 1/(5e^{cos(5x)-1)}+C