What is the integral of tan(θ)ln(sec(θ)) ?

The integral of tan(θ)ln(sec(θ)) is (ln^2(sec(θ)))/2+C

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integral of tan(θ)ln(sec(θ))

\int tan (θ) ln (sec (θ)) d θ

\frac{ln ^{2} ( sec ( θ ) )}{2} + C

Solution steps

\int tan (θ) ln (sec (θ)) d θ

Apply u-substitution

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

= \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{v ^{2}}{2}

Substitute back

= \frac{ln ^{2} ( sec ( θ ) )}{2}

Add a constant to the solution

= \frac{ln ^{2} ( sec ( θ ) )}{2} + C

\frac{d r}{d s} = e^{r - 7 s}

\int [3 sin (x) + (\frac{1}{2}) tan (x)] d x

f (x) = 6 - 5 x^{2}

4 x^{7}

y = x^{2} + 3 x^{2} ln (x)

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