Question

Integral of tan(x)*ln(cos(x))

Answer 1

Answer:

$\int {tan(x)*ln(cos(x))} \, dx = -\frac{ln^2cos(x)}{2} + c$

Explanation:

Given

$tan(x)*ln(cos(x))$

Required

Integrate

This is represented as:

$\int {tan(x)*ln(cos(x))} \, dx$

Let

$u =- ln(cos(x))$

So that:

$\frac{du}{dx} = \frac{sin(x)}{cos(x)}$

Make dx the subject

$dx = \frac{cos(x)}{sin(x)}du$

$\int {tan(x)*ln(cos(x))} \, dx$ becomes

$\int {-u *tan(x)*\frac{cos(x)}{sin(x)}du$

Express tan x as sin x/ cos x

$\int {-u *\frac{sin(x)}{cos(x)}*\frac{cos(x)}{sin(x)}du$

$\int {-u du$

$-\int {u du$

Apply power rule

$-\frac{u^{1+1}}{1+1} + c$

$-\frac{u^{2}}{2} + c$

Substitute $u =- ln(cos(x))$

$-\frac{ln^2cos(x)}{2} + c$

Hence:

$\int {tan(x)*ln(cos(x))} \, dx = -\frac{ln^2cos(x)}{2} + c$