Help me please? with indefinite integral ln(cosx)dx/cos^2x

Question

Samantha Adkison · Accepted Answer

#I=ln(cos(x))tan(x)+tan(x)-x+C#

Our goal is to resolve

#I=intln(cos(x))/cos^2(x)dx=intln(cos(x))sec^2(x)dx#

Make use of integration by parts.

#intudv=uv-intvdu#

Let #u=ln(cos(x))# and #dv=sec^2(x)dx#

Then #du=-tan(x)dx# and #v=tan(x)#

#I=ln(cos(x))tan(x)-inttan(x)(-tan(x))dx#

#=ln(cos(x))tan(x)+inttan^2(x)dx#

Next, figure out the integral.

#I_1=inttan^2(x)dx#

Use the identity #tan^2(x)=sec^2(x)-1#

#I_1=intsec^2(x)dx-int1dx=tan(x)-x+C#

Thus,

#I=ln(cos(x))tan(x)+tan(x)-x+C#

Ezekiel Alexander · Answer

undefined To solve the indefinite integral $ \int \frac{\ln(\cos x)}{\cos^2 x} \, dx $, we can use the substitution method. Let's set $ u = \cos x $ and find $ du $.

$$ \frac{du}{dx} = -\sin x $$
$$ du = -\sin x \, dx $$

Now, let's rewrite the integral in terms of $ u $:

$$ \int \frac{\ln u}{u^2} \, du $$

This integral can be solved using integration by parts. Let $ dv = \frac{1}{u^2} \, du $, then $ v = -\frac{1}{u} $.

$$ \int \frac{\ln u}{u^2} \, du = -\frac{\ln u}{u} - \int \left(-\frac{1}{u}\right) \left(-\frac{1}{u^2}\right) \, du $$

$$ = -\frac{\ln u}{u} + \int \frac{1}{u^3} \, du $$

$$ = -\frac{\ln u}{u} - \frac{1}{2u^2} + C $$

Finally, substituting back $ u = \cos x $, we get:

$$ \int \frac{\ln(\cos x)}{\cos^2 x} \, dx = -\frac{\ln(\cos x)}{\cos x} - \frac{1}{2\cos^2 x} + C $$