Integration Help

Aug 2008
15
0
I would really appreciate your help with 3 integration problems that I am completely unsure about.

The first is (lnx)^2 dx
The Second is e^x sin(3x) dx
The third is a definitive integral between pi and zero, sin^3(x) dx


Thank you!
 
Jul 2009
555
298
Zürich
I would really appreciate your help with 3 integration problems that I am completely unsure about.

The first is (lnx)^2 dx
Substitute \(\displaystyle u := \ln(x)\) to get \(\displaystyle \int u^2\cdot e^u\, du\), then do integration by parts twice, until you have reduced the factor \(\displaystyle u^2\) to a constant. Finally, do the backsubstitution.

The Second is e^x sin(3x) dx
Do integration by parts twice (differentiating the factor \(\displaystyle \sin(3x)\), until you get a constant multiple of the original integral). Then solve that equation for the original integral.

The third is a definitive integral between pi and zero, sin^3(x) dx
Consider that \(\displaystyle \sin^3(x)=\sin^2(x)\cdot\sin(x)=(1-\cos^2(x))\cdot\sin(x)\).
Thus substitute \(\displaystyle u := \cos(x)\), because \(\displaystyle \sin(x)\) is (almost) the derivative of \(\displaystyle u\).
 
Last edited:

Prove It

MHF Helper
Aug 2008
12,883
4,999
I would really appreciate your help with 3 integration problems that I am completely unsure about.

The first is (lnx)^2 dx
The Second is e^x sin(3x) dx
The third is a definitive integral between pi and zero, sin^3(x) dx


Thank you!
\(\displaystyle \int{(\ln{x})^2\,dx} = \int{\ln{x}\cdot\ln{x}\,dx}\).


Using integration by parts with

\(\displaystyle u = \ln{x}\) so that \(\displaystyle du = \frac{1}{x}\)

\(\displaystyle dv = \ln{x}\) so that \(\displaystyle v = x\ln{x} - x\)

we find

\(\displaystyle \int{\ln{x}\cdot\ln{x}\,dx} = \ln{x}(x\ln{x} - x) - \int{\frac{x\ln{x} - x}{x}\,dx}\)

\(\displaystyle =x(\ln{x})^2 - x\ln{x} - \int{\ln{x} - 1\,dx}\)

\(\displaystyle =x(\ln{x})^2 - x\ln{x} - (x\ln{x} - x - x) + C\)

\(\displaystyle = x(\ln{x})^2 - x\ln{x} - x\ln{x} + 2x + C\)

\(\displaystyle = x(\ln{x})^2 - 2x\ln{x} + 2x + C\).
 
Aug 2008
15
0
Help

Thank you for your help!
I have figured out the last two questions, but I am still a bit hazy over the definite integral part. I worked out that the normal integral for that question is

1/3 cos^3 (x)-cos(x)+c

Is that correct?
The definite integral between pi and zero still confuses me though. So, u=cos(x), and I think you plug in pi and zero into that, which gives you -1 and 1. What do I do then?

Thank you!
 
Sep 2008
1,261
539
West Malaysia
Thank you for your help!
I have figured out the last two questions, but I am still a bit hazy over the definite integral part. I worked out that the normal integral for that question is

1/3 cos^3 (x)-cos(x)+c

Is that correct?
The definite integral between pi and zero still confuses me though. So, u=cos(x), and I think you plug in pi and zero into that, which gives you -1 and 1. What do I do then?

Thank you!
you only 'modify' the limits when you 'modify' the integrand , ie make substitution etc ..

so basically its 1/3 u^3-u , so here you use the modified limits , 1 and -1

when you put it back to its original form , then you use the original limits .
 
Jul 2009
555
298
Zürich
Thank you for your help!
I have figured out the last two questions, but I am still a bit hazy over the definite integral part. I worked out that the normal integral for that question is

1/3 cos^3 (x)-cos(x)+c

Is that correct?
Yes.

The definite integral between pi and zero still confuses me though. So, u=cos(x), and I think you plug in pi and zero into that, which gives you -1 and 1. What do I do then?
Two possibilities, first, without backsubsitution you need to transform the limits of the integral along with the integrand:
\(\displaystyle \int_0^\pi \sin^3(x)\,dx=\int_{\cos(0)}^{\cos(\pi)}(u^2-1)du=\left[\frac{1}{3}u^3-u\right]_{u=\cos(0)}^{\cos(\pi)}=\left[\frac{1}{3}u^3-u\right]_{u=1}^{-1}=\ldots\)

Second, with backsubsitution:
\(\displaystyle \int_0^\pi\sin^3(x)\, dx=\left[\frac{1}{3}\cos^3(x)-\cos(x)\right]_{x=0}^\pi=\ldots\)
Don't get confused because of the constant of integration, c, because it does not matter anyway: it would cancel out in the final difference.