Integral of: tanxln(cosx)dx

[MarkSA]

Hello,

1) Find the integral of: tanxln(cosx)dx

Yikes. if I u-sub I don't seem to get anything useful. dx of tanx is sec^2x.. no. dx of cosx is -sinx.. no.

Any ideas?

Thanks

 

[PAULK]

Hello,

1) Find the integral of: tanxln(cosx)dx

Yikes. if I u-sub I don't seem to get anything useful. dx of tanx is sec^2x.. no. dx of cosx is -sinx.. no.

Any ideas?

Your integral is:

{ sin x dx
| ----------------
} cos x ln (cos x)

Try a t-substitution, same as a u-subst, but a different letter.

t = cos x , dt = - sin x dx.

{ - dt
| ---------
} t ln (t)

Now u = ln t, du = dt/dt

{ - du
| --------- = - ln u = - ln ln t = - ln ln cos x
} u

 

[galactus]

We can also try parts with this one.

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

Now, let $\displaystyle u=ln(cos(x)), \;\ dv=tan(x)dx, \;\ du = -tan(x)dx, \;\ v = -ln(cos(x))$

And we have:

$\displaystyle \int{tan(x)ln(cos(x))}dx = -(ln(cos(x)))^{2}-\int{ln(cos(x))tan(x)}dx$

Then, by adding $\displaystyle \int{tan(x)ln(cos(x))}dx$ to both sides, we have:

$\displaystyle \int{tan(x)ln(cos(x))}dx=\frac{-(ln(cos(x)))^{2}}{2}$

 

[soroban]

Hello, Mark!

There is a simple substitution . . . You just have to recognize it.

$\displaystyle 1)\;\text{Integrate: }\;\int \tan x\ln(\cos x)\,dx$

$\displaystyle \text{We have: }\;\int\ln(\cos x)\,\tan x\,dx$

$\displaystyle \text{Let: }\:u \:=\:\ln(\cos x)$

. . $\displaystyle du \:=\:\frac{-\sin x}{\cos x}\,dx \:=\:-\tan x\,dx\quad\Rightarrow\quad \tan x\,dx \:=\:-du$


$\displaystyle \text{Substitute: }\;\int u (-du) \;\;=\;\;-\int u\,du\;\;=\;\;-\frac{1}{2}u^2 + C$


$\displaystyle \text{Back-substitute: }\;\frac{1}{2}\left[\ln(\cos x)\right]^2 + C$