What is the integral of 9(tan(x))(ln(cos(x))) ?

The integral of 9(tan(x))(ln(cos(x))) is -9/2 ln^2(cos(x))+C

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integral of 9(tan(x))(ln(cos(x)))

\int 9 (tan (x)) (ln (cos (x))) d x

- \frac{9}{2} ln^{2} (cos (x)) + C

Solution steps

\int 9 tan (x) ln (cos (x)) d x

Take the constant out:

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

= 9 \cdot \int tan (x) ln (cos (x)) d x

Apply u-substitution

= 9 \cdot \int - \frac{ln ( u )}{u} d u

Take the constant out:

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

= 9 (- \int \frac{ln ( u )}{u} d u)

Apply u-substitution

= 9 (- \int v d v)

Apply the Power Rule

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

Substitute back

= 9 (- \frac{ln ^{2} ( cos ( x ) )}{2})

Simplify

9 (- \frac{ln ^{2} ( cos ( x ) )}{2}) : - \frac{9}{2} ln^{2} (cos (x))

= - \frac{9}{2} ln^{2} (cos (x))

Add a constant to the solution

= - \frac{9}{2} ln^{2} (cos (x)) + C

What is the integral of 9(tan(x))(ln(cos(x))) ?
The integral of 9(tan(x))(ln(cos(x))) is -9/2 ln^2(cos(x))+C