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

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]

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