What is the integral of (sin(x))/(cos(x)(cos(x)-1)) ?

The integral of (sin(x))/(cos(x)(cos(x)-1)) is ln|cos(x)|-ln|cos(x)-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(x))/(cos(x)(cos(x)-1))

\int \frac{sin ( x )}{cos ( x ) ( cos ( x ) - 1 )} d x

ln ∣ cos (x) ∣ - ln ∣ cos (x) - 1 ∣ + C

Solution steps

\int \frac{sin ( x )}{cos ( x ) ( cos ( x ) - 1 )} d x

Rewrite using trig identities

= \int \frac{tan ( x )}{cos ( x ) - 1} d x

Apply u-substitution

: \int - \frac{1}{u ( u - 1 )} d u

= \int - \frac{1}{u ( u - 1 )} d u

Take the constant out:

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

= - \int \frac{1}{u ( u - 1 )} d u

Take the partial fraction of

\frac{1}{u ( u - 1 )} : - \frac{1}{u} + \frac{1}{u - 1}

= - \int - \frac{1}{u} + \frac{1}{u - 1} d u

Apply the Sum Rule:

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

= - (- \int \frac{1}{u} d u + \int \frac{1}{u - 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= - (- ln ∣ u ∣ + ln ∣ u - 1 ∣)

Substitute back

u = cos (x)

= - (- ln ∣ cos (x) ∣ + ln ∣ cos (x) - 1 ∣)

Simplify

= ln ∣ cos (x) ∣ - ln ∣ cos (x) - 1 ∣

Add a constant to the solution

= ln ∣ cos (x) ∣ - ln ∣ cos (x) - 1 ∣ + C

\int - csc^{2} (3 x) d x

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

\int 1 0^{3} x^{2} d x

\int \frac{x ^{4}}{x ^{4} + 5 x ^{2} + 4} d x

\int cos (x - y) d x

What is the integral of (sin(x))/(cos(x)(cos(x)-1)) ?
The integral of (sin(x))/(cos(x)(cos(x)-1)) is ln|cos(x)|-ln|cos(x)-1|+C