What is the integral of 9tan(3x)ln(sec(3x)) ?

The integral of 9tan(3x)ln(sec(3x)) is 3/2 ln^2(sec(3x))+C

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integral of 9tan(3x)ln(sec(3x))

\int 9 tan (3 x) ln (sec (3 x)) d x

\frac{3}{2} ln^{2} (sec (3 x)) + C

Solution steps

\int 9 tan (3 x) ln (sec (3 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 (3 x) ln (sec (3 x)) d x

Apply u-substitution

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

= 9 \cdot \int tan (u) ln (sec (u)) \frac{1}{3} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int tan (u) ln (sec (u)) d u

Apply u-substitution

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

= 9 \cdot \frac{1}{3} \cdot \int \frac{ln ( v )}{v} d v

Apply u-substitution

: \int w d w

= 9 \cdot \frac{1}{3} \cdot \int w d w

Apply the Power Rule

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

= 9 \cdot \frac{1}{3} \cdot \frac{w ^{2}}{2}

Substitute back

= 9 \cdot \frac{1}{3} \cdot \frac{ln ^{2} ( sec ( 3 x ) )}{2}

Simplify

9 \cdot \frac{1}{3} \cdot \frac{ln ^{2} ( sec ( 3 x ) )}{2} : \frac{3}{2} ln^{2} (sec (3 x))

= \frac{3}{2} ln^{2} (sec (3 x))

Add a constant to the solution

= \frac{3}{2} ln^{2} (sec (3 x)) + C

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What is the integral of 9tan(3x)ln(sec(3x)) ?
The integral of 9tan(3x)ln(sec(3x)) is 3/2 ln^2(sec(3x))+C