What is the integral of 5tan(2x)ln(sec(2x)) ?

The integral of 5tan(2x)ln(sec(2x)) is 5/4 ln^2(sec(2x))+C

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integral of 5tan(2x)ln(sec(2x))

\int 5 tan (2 x) ln (sec (2 x)) d x

\frac{5}{4} ln^{2} (sec (2 x)) + C

Solution steps

\int 5 tan (2 x) ln (sec (2 x)) d x

Take the constant out:

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

= 5 \cdot \int tan (2 x) ln (sec (2 x)) d x

Apply u-substitution

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

= 5 \cdot \int tan (u) ln (sec (u)) \frac{1}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int tan (u) ln (sec (u)) d u

Apply u-substitution

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{ln ( v )}{v} d v

Apply u-substitution

: \int w d w

= 5 \cdot \frac{1}{2} \cdot \int w d w

Apply the Power Rule

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

= 5 \cdot \frac{1}{2} \cdot \frac{w ^{2}}{2}

Substitute back

= 5 \cdot \frac{1}{2} \cdot \frac{ln ^{2} ( sec ( 2 x ) )}{2}

Simplify

5 \cdot \frac{1}{2} \cdot \frac{ln ^{2} ( sec ( 2 x ) )}{2} : \frac{5}{4} ln^{2} (sec (2 x))

= \frac{5}{4} ln^{2} (sec (2 x))

Add a constant to the solution

= \frac{5}{4} ln^{2} (sec (2 x)) + C

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What is the integral of 5tan(2x)ln(sec(2x)) ?
The integral of 5tan(2x)ln(sec(2x)) is 5/4 ln^2(sec(2x))+C