What is the integral of 8tan(2x)ln(cos(2x)) ?

The integral of 8tan(2x)ln(cos(2x)) is -2ln^2(cos(2x))+C

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integral of 8tan(2x)ln(cos(2x))

\int 8 tan (2 x) ln (cos (2 x)) d x

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

Solution steps

\int 8 tan (2 x) ln (cos (2 x)) d x

Take the constant out:

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

= 8 \cdot \int tan (2 x) ln (cos (2 x)) d x

Apply u-substitution

: \int tan (u) ln (cos (u)) \frac{1}{2} d u

= 8 \cdot \int tan (u) ln (cos (u)) \frac{1}{2} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int tan (u) ln (cos (u)) d u

Apply u-substitution

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

= 8 \cdot \frac{1}{2} \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

= 8 \cdot \frac{1}{2} (- \int \frac{ln ( v )}{v} d v)

Apply u-substitution

: \int w d w

= 8 \cdot \frac{1}{2} (- \int w d w)

Apply the Power Rule

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

= 8 \cdot \frac{1}{2} (- \frac{w ^{2}}{2})

Substitute back

= 8 \cdot \frac{1}{2} (- \frac{ln ^{2} ( cos ( 2 x ) )}{2})

Simplify

8 \cdot \frac{1}{2} (- \frac{ln ^{2} ( cos ( 2 x ) )}{2}) : - 2 ln^{2} (cos (2 x))

= - 2 ln^{2} (cos (2 x))

Add a constant to the solution

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

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What is the integral of 8tan(2x)ln(cos(2x)) ?
The integral of 8tan(2x)ln(cos(2x)) is -2ln^2(cos(2x))+C