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Integral of: tanxln(cosx)dx
- Thread starter MarkSA
- Start date
Your integral is:MarkSA said: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?
{ 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
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:
\(\displaystyle \int{tan(x)ln(cos(x))}dx=\frac{-(ln(cos(x)))^{2}}{2}\)
\(\displaystyle \int{tan(x)ln(cos(x))}dx\)
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:
\(\displaystyle \int{tan(x)ln(cos(x))}dx=\frac{-(ln(cos(x)))^{2}}{2}\)
Hello, Mark!
There is a simple substitution . . . You just have to recognize it.
\(\displaystyle \text{We have: }\;\int\ln(\cos x)\,\tan x\,dx\)
\(\displaystyle \text{Let: }\:u \:=\:\ln(\cos x)\)
. . \(\displaystyle du \:=\:\frac{-\sin x}{\cos x}\,dx \:=\:-\tan x\,dx\quad\Rightarrow\quad \tan x\,dx \:=\:-du\)
\(\displaystyle \text{Substitute: }\;\int u (-du) \;\;=\;\;-\int u\,du\;\;=\;\;-\frac{1}{2}u^2 + C\)
\(\displaystyle \text{Back-substitute: }\;\frac{1}{2}\left[\ln(\cos x)\right]^2 + C\)
There is a simple substitution . . . You just have to recognize it.
\(\displaystyle 1)\;\text{Integrate: }\;\int \tan x\ln(\cos x)\,dx\)
\(\displaystyle \text{We have: }\;\int\ln(\cos x)\,\tan x\,dx\)
\(\displaystyle \text{Let: }\:u \:=\:\ln(\cos x)\)
. . \(\displaystyle du \:=\:\frac{-\sin x}{\cos x}\,dx \:=\:-\tan x\,dx\quad\Rightarrow\quad \tan x\,dx \:=\:-du\)
\(\displaystyle \text{Substitute: }\;\int u (-du) \;\;=\;\;-\int u\,du\;\;=\;\;-\frac{1}{2}u^2 + C\)
\(\displaystyle \text{Back-substitute: }\;\frac{1}{2}\left[\ln(\cos x)\right]^2 + C\)