What is the integral of (cos(x))/(8sin^2(x)+9sin(x)) ?

The integral of (cos(x))/(8sin^2(x)+9sin(x)) is -1/9 (ln| 16/9 sin(x)+2|-ln(16/9 |sin(x)|))+C

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integral of (cos(x))/(8sin^2(x)+9sin(x))

\int \frac{cos ( x )}{8 sin ^{2} ( x ) + 9 sin ( x )} d x

- \frac{1}{9} (ln \frac{16}{9} sin (x) + 2 - ln (\frac{16}{9} ∣ sin (x) ∣)) + C

Solution steps

\int \frac{cos ( x )}{8 sin ^{2} ( x ) + 9 sin ( x )} d x

Apply u-substitution

: \int \frac{1}{8 u ^{2} + 9 u} d u

= \int \frac{1}{8 u ^{2} + 9 u} d u

Complete the square

8 u^{2} + 9 u : 8 (u + \frac{9}{16})^{2} - \frac{81}{32}

= \int \frac{1}{8 ( u + \frac{9}{16} ) ^{2} - \frac{81}{32}} d u

Apply u-substitution

: \int \frac{32}{256 v ^{2} - 81} d v

= \int \frac{32}{256 v ^{2} - 81} d v

Take the constant out:

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

= 32 \cdot \int \frac{1}{256 v ^{2} - 81} d v

Apply Integral Substitution

= 32 \cdot \int \frac{1}{144 ( w ^{2} - 1 )} d w

Take the constant out:

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

= 32 \cdot \frac{1}{144} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 32 \cdot \frac{1}{144} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

Take the constant out:

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

= 32 \cdot \frac{1}{144} (- \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

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

= 32 \cdot \frac{1}{144} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 32 \cdot \frac{1}{144} (- (\frac{ln \frac{16}{9} ( sin ( x ) + \frac{9}{16} ) + 1}{2} - \frac{ln \frac{16}{9} ( sin ( x ) + \frac{9}{16} ) - 1}{2}))

Simplify

32 \cdot \frac{1}{144} (- (\frac{ln \frac{16}{9} ( sin ( x ) + \frac{9}{16} ) + 1}{2} - \frac{ln \frac{16}{9} ( sin ( x ) + \frac{9}{16} ) - 1}{2})) : - \frac{1}{9} (ln \frac{16}{9} sin (x) + 2 - ln (\frac{16}{9} ∣ sin (x) ∣))

= - \frac{1}{9} (ln \frac{16}{9} sin (x) + 2 - ln (\frac{16}{9} ∣ sin (x) ∣))

Add a constant to the solution

= - \frac{1}{9} (ln \frac{16}{9} sin (x) + 2 - ln (\frac{16}{9} ∣ sin (x) ∣)) + C

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What is the integral of (cos(x))/(8sin^2(x)+9sin(x)) ?
The integral of (cos(x))/(8sin^2(x)+9sin(x)) is -1/9 (ln| 16/9 sin(x)+2|-ln(16/9 |sin(x)|))+C