- #1
paulmdrdo1
- 385
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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,paulmdrdo said:what technique should I use here?
∫dx/1+e^x
∫csc^4x dx/cot^2x
∫cos^7 2x dx
Petrus said: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,paulmdrdo said: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.
ZaidAlyafey said: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 ?
paulmdrdo said: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(Clapping)(I never think like that...)ZaidAlyafey said:\(\displaystyle \frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}\)
your third and fourth step I did not understandZaidAlyafey said:\(\displaystyle \Large \frac{\csc^4x}{\cot^2x} = \frac{\frac{1}{\sin^4 x}}{\frac{\cos^2 x}{\sin^2 x}} = \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x} = \sec^2 x + \csc^2 x \)
Petrus said:\(\displaystyle \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x}\)
How does that work?
Hey Zaid,ZaidAlyafey said:\(\displaystyle 1= \sin^2 x + \cos^2 x \)
I substituted instead of \(\displaystyle 1\) in the numerator the term \(\displaystyle \sin^2 x + \cos^2 x \).
To solve ∫dx/1+e^x, we can use the substitution method by letting u = 1+e^x. This will result in du = e^x dx, which we can then substitute into the original integral. The integral then becomes ∫du/u, which can be easily solved using the natural log function.
The most efficient approach for solving ∫csc^4x/cot^2x is to use the trigonometric identity csc^2x = 1 + cot^2x. This will result in the integral becoming ∫(1+cot^2x)/cot^2x, which can then be simplified and solved using integration techniques.
Yes, a u-substitution can be used to solve ∫cos^7 2x. Let u = 2x, which results in du = 2 dx. We can then substitute this into the integral, which becomes (1/2)∫cos^7 u du. From here, we can use the power reduction formula for cosine to simplify and solve the integral.
Yes, there are various formulas and rules that can be used to solve integrals involving trigonometric functions. Some of the most commonly used ones include the power reduction formulas, integration by parts, and trigonometric identities such as csc^2x = 1 + cot^2x and sec^2x = 1 + tan^2x.
One tip for solving integrals involving both exponential and trigonometric functions is to look for a substitution that can simplify the integral. For example, if the integral contains e^x and sin x, we can use the substitution u = e^x to eliminate the exponential term and make the integral easier to solve.