What is the integral of 7tan(3x)ln(cos(3x)) ?

The integral of 7tan(3x)ln(cos(3x)) is -7/6 ln^2(cos(3x))+C

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integral of 7tan(3x)ln(cos(3x))

\int 7 tan (3 x) ln (cos (3 x)) d x

- \frac{7}{6} ln^{2} (cos (3 x)) + C

Solution steps

\int 7 tan (3 x) ln (cos (3 x)) d x

Take the constant out:

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

= 7 \cdot \int tan (3 x) ln (cos (3 x)) d x

Apply u-substitution

: \int tan (u) ln (cos (u)) \frac{1}{3} d u

= 7 \cdot \int tan (u) ln (cos (u)) \frac{1}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int tan (u) ln (cos (u)) d u

Apply u-substitution

: \int - \frac{ln ( v )}{v} d v

= 7 \cdot \frac{1}{3} \cdot \int - \frac{ln ( v )}{v} d v

Take the constant out:

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

= 7 \cdot \frac{1}{3} (- \int \frac{ln ( v )}{v} d v)

Apply u-substitution

: \int w d w

= 7 \cdot \frac{1}{3} (- \int w d w)

Apply the Power Rule

: \frac{w ^{2}}{2}

= 7 \cdot \frac{1}{3} (- \frac{w ^{2}}{2})

Substitute back

= 7 \cdot \frac{1}{3} (- \frac{ln ^{2} ( cos ( 3 x ) )}{2})

Simplify

7 \cdot \frac{1}{3} (- \frac{ln ^{2} ( cos ( 3 x ) )}{2}) : - \frac{7}{6} ln^{2} (cos (3 x))

= - \frac{7}{6} ln^{2} (cos (3 x))

Add a constant to the solution

= - \frac{7}{6} ln^{2} (cos (3 x)) + C

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What is the integral of 7tan(3x)ln(cos(3x)) ?
The integral of 7tan(3x)ln(cos(3x)) is -7/6 ln^2(cos(3x))+C