Integrating Tricky Functions: Problems with Homework Statements

[La_Lune]

Homework Statement

a) ∫xe^x/√(1+e^x) dx from ln3 to ln8
b)∫arccos(tanx)dx from -π/4 to π/4

Homework Equations

uv-∫vdu? Not so sure about this

The Attempt at a Solution

For question a I tried to make e^x/√(1+e^x)=dv and x=u, but then my result was different from the correct answer, so I guess perhaps I was proceeding the wrong way...For question b I really don't have any idea, perhaps I should substitute tanx with something? Thanks for your help!

 

[SammyS]

For part(a), that looks like your method should work

dv = e^x/√(1+e^x) dx → v = 2√(1+e^x)

and of course, u = x → du = dx

For part (b), all I could come up with is a graphical solution combined with relating the integral of a function with the integral of the function's inverse.

Added in edit:

cos(θ) = sin(π/2 - θ)

So, arccos(u) = π/2 - arcsin(u)

Therefore, $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arccos(\tan(x))dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{\pi}{2}-\arcsin(\tan(x))\right)dx\ .$$

Split this into two integrals, and notice that arcsin(tan(x)) is and odd function.

 

[La_Lune]

For part(a), that looks like your method should work

dv = e^x/√(1+e^x) dx → v = 2√(1+e^x)

and of course, u = x → du = dx

For part (b), all I could come up with is a graphical solution combined with relating the integral of a function with the integral of the function's inverse.

Added in edit:

cos(θ) = sin(π/2 - θ)

So, arccos(u) = π/2 - arcsin(u)

Therefore, $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arccos(\tan(x))dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{\pi}{2}-\arcsin(\tan(x))\right)dx\ .$$

Split this into two integrals, and notice that arcsin(tan(x)) is and odd function.

Thank you so much SammyS! The method for question b worked like a charm! Can you please tell me how I shall proceed with question a? I mean using integration by parts I got this:

2x√(1+e^x)-2∫√(1+e^x)dx but I don't know how to find the integral of the second part...

 

[SammyS]

Thank you so much SammyS! The method for question b worked like a charm! Can you please tell me how I shall proceed with question a? I mean using integration by parts I got this:

2x√(1+e^x)-2∫√(1+e^x)dx but I don't know how to find the integral of the second part...

That integral looks rather tough. WolframAlpha does a rather involved set of substitutions, but try the following instead.

$$\text{Let }u = \sqrt{1 + e^x}\ \ \to\ \ x=\ln(u^2-1)=\ln(u+1)+\ln(u-1)\ \ \to\ \ dx=\left(\frac{1}{u+1}+\frac{1}{u-1}\right)du\ .$$

This works out fairly well.

 

[La_Lune]

That integral looks rather tough. WolframAlpha does a rather involved set of substitutions, but try the following instead.

$$\text{Let }u = \sqrt{1 + e^x}\ \ \to\ \ x=\ln(u^2-1)=\ln(u+1)+\ln(u-1)\ \ \to\ \ dx=\left(\frac{1}{u+1}+\frac{1}{u-1}\right)du\ .$$

This works out fairly well.

I thank you so much~it seems that one should really be flexible while trying to use substitution :)