What is the integral of-tan(2x+1)ln(cos(2x+1)) ?

The integral of-tan(2x+1)ln(cos(2x+1)) is 1/4 ln^2(cos(2x+1))+C

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integral of-tan(2x+1)ln(cos(2x+1))

\int - tan (2 x + 1) ln (cos (2 x + 1)) d x

\frac{1}{4} ln^{2} (cos (2 x + 1)) + C

Solution steps

\int - tan (2 x + 1) ln (cos (2 x + 1)) d x

Take the constant out:

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

= - \int tan (2 x + 1) ln (cos (2 x + 1)) d x

Apply u-substitution

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

= - \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

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

Apply u-substitution

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

= - \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

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

Apply u-substitution

: \int w d w

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

- \frac{1}{2} (- \frac{ln ^{2} ( cos ( 2 x + 1 ) )}{2}) : \frac{1}{4} ln^{2} (cos (2 x + 1))

= \frac{1}{4} ln^{2} (cos (2 x + 1))

Add a constant to the solution

= \frac{1}{4} ln^{2} (cos (2 x + 1)) + C

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What is the integral of-tan(2x+1)ln(cos(2x+1)) ?
The integral of-tan(2x+1)ln(cos(2x+1)) is 1/4 ln^2(cos(2x+1))+C