1. ## More Intergration questions

Hi

The following question i am having trouble solving:

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

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

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

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

$\frac{x}{4} - \frac{2sinx}{4} + c$

$\frac{x}{4} - \frac{sinx}{2} + c$

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

$u = e^{-x} du = e^{-x}$

$dv = x v = \frac{x^2}{2}$

$\int e^{-x}x = e^{-x} * \frac{x^2}{2} - \int \frac{x^2}{2} * e^{-x}$

$=\frac{x^2e^{-x}}{2} - \frac{1}{2} \int -x^{2}e^{-x}$

$=\frac{x^2e^{-x}}{2} - \frac{1}{2} [ -x^{2}e^{-x} - \int -e^{-x} * 2x]$

eventually i get

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

P.S

2. Originally Posted by Paymemoney
Hi

The following question i am having trouble solving:

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

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

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

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

$\frac{x}{4} - \frac{2sinx}{4} + c$

$\frac{x}{4} - \frac{sinx}{2} + c$

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

$u = e^{-x} du = e^{-x}$

$dv = x v = \frac{x^2}{2}$

$\int e^{-x}x = e^{-x} * \frac{x^2}{2} - \int \frac{x^2}{2} * e^{-x}$

$=\frac{x^2e^{-x}}{2} - \frac{1}{2} \int -x^{2}e^{-x}$

$=\frac{x^2e^{-x}}{2} - \frac{1}{2} [ -x^{2}e^{-x} - \int -e^{-x} * 2x]$

eventually i get

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

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

fix that.

$\int xe^{-x} \, dx$

$u = x$ ... $dv = e^{-x} \, dx$

$du = dx$ ... $v = -e^{-x}$

$\int xe^{-x} \, dx = -xe^{-x} + \int e^{-x} \, dx$

$\int xe^{-x} \, dx = -xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C$

3. ## Gotnomoney

Hi Pay me! see if the attachment helps.

Be well,
T

[IMG]file:///C:/Users/Terry/AppData/Local/Temp/moz-screenshot-3.png[/IMG]

4. $\int xe^{-x} \, dx = -xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C$[/quote]

5. Originally Posted by boardguy67
$\int xe^{-x} \, dx = -xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C$
i believe that is correct, because i worked it out again and i got the same answer.

6. Would this be correct?

$\int cos^4(x)$

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

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

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

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

7. ## 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

8. Originally Posted by Paymemoney
Would this be correct?

$\int cos^4(x)$

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

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

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

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

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

Just finish it off from here

9. Using the substitution method

$u = cos(2x) du = -2sin(2x)$

$=\frac{1}{4} \int x + \frac{2u}{2} + \frac{u^3}{3} du$

$=\frac{x}{4} + \frac{2u^2}{8} + \frac{u^3}{12} + c$

$=\frac{x}{4} + \frac{u^2}{4} + \frac{u^3}{12} + c$

$=\frac{3x}{12} + \frac{3u^2}{12} + \frac{u^3}{12} + c$

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

10. Since $\cos{2x} = \cos^2{x} - \sin^2{x}$

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

$\cos{2x} = 2\cos^2{x} - 1$

$\cos{2x} + 1 = 2\cos^2{x}$

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

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

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

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

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

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

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

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

$= \frac{1}{8}\cos{4x} + \frac{1}{2}\cos{2x} + \frac{3}{8}$.

Therefore

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

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

11. ## Reduction formula

Sorry friend, I snoozed a bit, and am gonna snooze more in a sec. For simplicity, the attachment is the maple output for your problem. It's a bit...shall we say...longish, but it should work for you. Son't let the filename throw you, it's your exercise.

Be well
t

12. thanks guys