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paulmdrdo
Active member
- May 13, 2013
- 386
what technique should I use here?
∫dx/1+e^x
∫csc^4x dx/cot^2x
∫cos^7 2x dx
∫dx/1+e^x
∫csc^4x dx/cot^2x
∫cos^7 2x dx
Hello Paul,what technique should I use here?
∫dx/1+e^x
∫csc^4x dx/cot^2x
∫cos^7 2x dx
is this correct?Hello Paul,
For the first one I asume you mean
\(\displaystyle \int\frac{dx}{1+e^x}\)
start with multiply both side with \(\displaystyle e^x\) so we got:
\(\displaystyle \int\frac{e^x}{e^x(1+e^x)}dx\)
subsitute \(\displaystyle u=e^x+1 <=> du=e^xdx\)
so our integrate become
\(\displaystyle \int \frac{du}{(u-1)u}\)
I leave the partial fraction to you
For the rest I need to think more...
Regards,
\(\displaystyle |\pi\rangle\)
Hello Paul,is this correct?
1/(u*(u-1)) = 1/(u-1) + 1/u
ln(u-1) + ln(u) + c
= ln(e^x) + ln(e^x+1) + c
= x + ln(e^x+1) + c.
i have a feeling that this correct. is it?You missed a sign in the partial fraction .
Here is an another way to do it
\(\displaystyle \frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}\)
Now you realize that the numerator is the derivative of the denominator , can you finish now ?
Correct !i have a feeling that this correct. is it?
=>∫(e^x+1)/e^x+1) dx - ∫e^x dx /(e^x+1)
=>∫dx - ∫d(e^x+1)/e^x+1
=>x - ln(e^x+1) + c
Amazing\(\displaystyle \frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}\)
your third and fourth step I did not understand\(\displaystyle \Large \frac{\csc^4x}{\cot^2x} = \frac{\frac{1}{\sin^4 x}}{\frac{\cos^2 x}{\sin^2 x}} = \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x} = \sec^2 x + \csc^2 x \)
\(\displaystyle 1= \sin^2 x + \cos^2 x \)\(\displaystyle \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x}\)
How does that work?
Hey Zaid,\(\displaystyle 1= \sin^2 x + \cos^2 x \)
I substituted instead of \(\displaystyle 1\) in the numerator the term \(\displaystyle \sin^2 x + \cos^2 x \).