What is the integral of (cos(x))/(sin^2(x)-81) ?

The integral of (cos(x))/(sin^2(x)-81) is -1/18 (ln|(sin(x))/9+1|-ln|(sin(x))/9-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cos(x))/(sin^2(x)-81)

\int \frac{cos ( x )}{sin ^{2} ( x ) - 81} d x

- \frac{1}{18} (ln \frac{sin ( x )}{9} + 1 - ln \frac{sin ( x )}{9} - 1) + C

Solution steps

\int \frac{cos ( x )}{sin ^{2} ( x ) - 81} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 81} d u

= \int \frac{1}{u ^{2} - 81} d u

Apply Integral Substitution

= \int \frac{1}{9 ( v ^{2} - 1 )} d v

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= \frac{1}{9} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{9} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{9} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= \frac{1}{9} - \frac{ln \frac{s i n ( x )}{9} + 1}{2} - \frac{ln \frac{s i n ( x )}{9} - 1}{2}

Simplify

\frac{1}{9} - \frac{ln \frac{s i n ( x )}{9} + 1}{2} - \frac{ln \frac{s i n ( x )}{9} - 1}{2} : - \frac{1}{18} (ln \frac{sin ( x )}{9} + 1 - ln \frac{sin ( x )}{9} - 1)

= - \frac{1}{18} (ln \frac{sin ( x )}{9} + 1 - ln \frac{sin ( x )}{9} - 1)

Add a constant to the solution

= - \frac{1}{18} (ln \frac{sin ( x )}{9} + 1 - ln \frac{sin ( x )}{9} - 1) + C

x \to 0 + lim (e^{x})

\frac{d}{d x} (ln (cos (e^{x^{2} - s e c (x)})))

\int \frac{9 e ^{- \frac{1}{x}}}{x ^{2}} d x

f (x) = x^{2} + 1, at x = 2

What is the integral of (cos(x))/(sin^2(x)-81) ?
The integral of (cos(x))/(sin^2(x)-81) is -1/18 (ln|(sin(x))/9+1|-ln|(sin(x))/9-1|)+C