More Intergration questions

Dec 2008
509
2
Hi

The following question i am having trouble solving:

1)Find \(\displaystyle \int cos^4(x) dx\)
This is what i have done

\(\displaystyle \frac{1}{4} \int (1+ cos(2x))\)

\(\displaystyle \frac{1}{4} \int (1+ cos(2x))\)

\(\displaystyle \frac(x - \frac{sin(2x)}{2}\)

\(\displaystyle \frac{x}{4} - \frac{2sinx}{4} + c\)

\(\displaystyle \frac{x}{4} - \frac{sinx}{2} + c\)


2)Find \(\displaystyle \int xe^{-x} dx\)
This is what i have done

\(\displaystyle u = e^{-x} du = e^{-x}\)

\(\displaystyle dv = x v = \frac{x^2}{2}\)

\(\displaystyle \int e^{-x}x = e^{-x} * \frac{x^2}{2} - \int \frac{x^2}{2} * e^{-x}\)

\(\displaystyle =\frac{x^2e^{-x}}{2} - \frac{1}{2} \int -x^{2}e^{-x}\)

\(\displaystyle =\frac{x^2e^{-x}}{2} - \frac{1}{2} [ -x^{2}e^{-x} - \int -e^{-x} * 2x]\)

eventually i get

\(\displaystyle = x^{2}e^{-x} + xe^{-x} = e^{-x}(x+1) +c\)

P.S
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
Hi

The following question i am having trouble solving:

1)Find \(\displaystyle \int cos^4(x) dx\)
This is what i have done

\(\displaystyle \frac{1}{4} \int (1+ cos(2x))\)

\(\displaystyle \frac{1}{4} \int (1+ cos(2x))\)

\(\displaystyle \frac(x - \frac{sin(2x)}{2}\)

\(\displaystyle \frac{x}{4} - \frac{2sinx}{4} + c\)

\(\displaystyle \frac{x}{4} - \frac{sinx}{2} + c\)


2)Find \(\displaystyle \int xe^{-x} dx\)
This is what i have done

\(\displaystyle u = e^{-x} du = e^{-x}\)

\(\displaystyle dv = x v = \frac{x^2}{2}\)

\(\displaystyle \int e^{-x}x = e^{-x} * \frac{x^2}{2} - \int \frac{x^2}{2} * e^{-x}\)

\(\displaystyle =\frac{x^2e^{-x}}{2} - \frac{1}{2} \int -x^{2}e^{-x}\)

\(\displaystyle =\frac{x^2e^{-x}}{2} - \frac{1}{2} [ -x^{2}e^{-x} - \int -e^{-x} * 2x]\)

eventually i get

\(\displaystyle = x^{2}e^{-x} + xe^{-x} = e^{-x}(x+1) +c\)

P.S
\(\displaystyle \cos^4{x} = \frac{[1+\cos(2x)]^{\textcolor{red}{2}}}{4}
\)

fix that.


\(\displaystyle \int xe^{-x} \, dx\)

\(\displaystyle u = x\) ... \(\displaystyle dv = e^{-x} \, dx\)

\(\displaystyle du = dx\) ... \(\displaystyle v = -e^{-x}\)

\(\displaystyle \int xe^{-x} \, dx = -xe^{-x} + \int e^{-x} \, dx\)

\(\displaystyle \int xe^{-x} \, dx = -xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C\)
 
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Reactions: Paymemoney
Apr 2010
17
2
Bend Oregon
\(\displaystyle \int xe^{-x} \, dx = -xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C\)[/quote]

Sure about this skeeter(assuming i cut the quote out right :))?
 
Dec 2008
509
2
\(\displaystyle \int xe^{-x} \, dx = -xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C\)
Sure about this skeeter(assuming i cut the quote out right :))?
i believe that is correct, because i worked it out again and i got the same answer.
 
Dec 2008
509
2
Would this be correct?

\(\displaystyle \int cos^4(x)\)

\(\displaystyle =\frac{1}{4} \int (1+cos(2x))^2
\)
\(\displaystyle =\frac{1}{4} \int \frac{(1+cos(2x))^3}{3} * -2sin(2x)\)

\(\displaystyle =\frac{1}{4} * \frac{-2sin(2x)(1+cos(2x))^3}{3}\)

\(\displaystyle =\frac{-2sin(2x)(1+cos(2x))^3}{12} = \frac{-sin(2x)(1+cos(2x))^3}{6}\)
 
Apr 2010
17
2
Bend Oregon
You have part of the idea,

First of all--Skeeter was right about the exponential integral. Sorry to doubt you skeeter...had the wrong function goin there for a minute. :(

As for your 4th power cos function, payme, use the reduction formula for cos to the n first, with n = 4, that leaves you with a mess in front of integral cos squared. Then use the reduction formula for integral cos squared on the last bit.

Try it one more time, then ill post my solution if you can't get there.

Be well,
T
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Would this be correct?

\(\displaystyle \int cos^4(x)\)

\(\displaystyle =\frac{1}{4} \int (1+cos(2x))^2
\)
\(\displaystyle =\frac{1}{4} \int \frac{(1+cos(2x))^3}{3} * -2sin(2x)\)

\(\displaystyle =\frac{1}{4} * \frac{-2sin(2x)(1+cos(2x))^3}{3}\)

\(\displaystyle =\frac{-2sin(2x)(1+cos(2x))^3}{12} = \frac{-sin(2x)(1+cos(2x))^3}{6}\)

\(\displaystyle \int cos^4(x)dx=\frac{1}{4}\int (1+cos(2x))^2dx=\frac{1}{4}\int (1+2cos(2x)+cos^2(2x))dx\)
\(\displaystyle =\frac{1}{4}\int dx +\frac{1}{4}\int cos(2x)(2dx)+\frac{1}{4}\int cos^2(2x)dx\)

Just finish it off from here
 
Dec 2008
509
2
Using the substitution method

\(\displaystyle u = cos(2x) du = -2sin(2x)\)

\(\displaystyle =\frac{1}{4} \int x + \frac{2u}{2} + \frac{u^3}{3} du\)

\(\displaystyle =\frac{x}{4} + \frac{2u^2}{8} + \frac{u^3}{12} + c\)

\(\displaystyle =\frac{x}{4} + \frac{u^2}{4} + \frac{u^3}{12} + c\)

\(\displaystyle =\frac{3x}{12} + \frac{3u^2}{12} + \frac{u^3}{12} + c\)


\(\displaystyle =\frac{3x}{12} + \frac{3cos^2(2x)}{12} + \frac{cos(2x)^3}{12} + c\)
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Since \(\displaystyle \cos{2x} = \cos^2{x} - \sin^2{x}\)

\(\displaystyle \cos{2x} = \cos^2{x} - (1 - \cos^2{x})\)

\(\displaystyle \cos{2x} = 2\cos^2{x} - 1\)

\(\displaystyle \cos{2x} + 1 = 2\cos^2{x}\)

\(\displaystyle \cos^2{x} = \frac{1}{2}(\cos{2x} + 1)\).



So \(\displaystyle \cos^4{x} = (\cos^2{x})^2\)

\(\displaystyle = \left[\frac{1}{2}(\cos{2x} + 1)\right]^2\)

\(\displaystyle = \frac{1}{4}(\cos{2x} + 1)^2\)

\(\displaystyle = \frac{1}{4}(\cos^2{2x} + 2\cos{2x} + 1)\)

\(\displaystyle = \frac{1}{4}\left[\frac{1}{2}(\cos{4x} + 1) + 2\cos{2x} + 1\right]\)

\(\displaystyle = \frac{1}{4}\left(\frac{1}{2}\cos{4x} + \frac{1}{2} + 2\cos{2x} + 1\right)\)

\(\displaystyle = \frac{1}{4}\left(\frac{1}{2}\cos{4x} + 2\cos{2x} + \frac{3}{2}\right)\)

\(\displaystyle = \frac{1}{8}\cos{4x} + \frac{1}{2}\cos{2x} + \frac{3}{8}\).


Therefore

\(\displaystyle \int{\cos^4{x}\,dx} = \int{\frac{1}{8}\cos{4x} + \frac{1}{2}\cos{2x} + \frac{3}{8}\,dx}\)

\(\displaystyle = \frac{1}{32}\sin{4x} + \frac{1}{4}\sin{2x} + \frac{3}{8}x + C\).