Answer:
[tex]\int {tan(x)*ln(cos(x))} \, dx = -\frac{ln^2cos(x)}{2} + c[/tex]
Explanation:
Given
[tex]tan(x)*ln(cos(x))[/tex]
Required
Integrate
This is represented as:
[tex]\int {tan(x)*ln(cos(x))} \, dx[/tex]
Let
[tex]u =- ln(cos(x))[/tex]
So that:
[tex]\frac{du}{dx} = \frac{sin(x)}{cos(x)}[/tex]
Make dx the subject
[tex]dx = \frac{cos(x)}{sin(x)}du[/tex]
[tex]\int {tan(x)*ln(cos(x))} \, dx[/tex] becomes
[tex]\int {-u *tan(x)*\frac{cos(x)}{sin(x)}du[/tex]
Express tan x as sin x/ cos x
[tex]\int {-u *\frac{sin(x)}{cos(x)}*\frac{cos(x)}{sin(x)}du[/tex]
[tex]\int {-u du[/tex]
[tex]-\int {u du[/tex]
Apply power rule
[tex]-\frac{u^{1+1}}{1+1} + c[/tex]
[tex]-\frac{u^{2}}{2} + c[/tex]
Substitute [tex]u =- ln(cos(x))[/tex]
[tex]-\frac{ln^2cos(x)}{2} + c[/tex]
Hence:
[tex]\int {tan(x)*ln(cos(x))} \, dx = -\frac{ln^2cos(x)}{2} + c[/tex]
