# More Intergration questions

#### Paymemoney

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
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.

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

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

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

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

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

Paymemoney

#### boardguy67

Gotnomoney

Hi Pay me! see if the attachment helps.

Be well,
T

#### Attachments

• 48 KB Views: 13

#### boardguy67

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

#### Paymemoney

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

#### Paymemoney

Would this be correct?

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

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

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

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

#### boardguy67

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
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

#### Paymemoney

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
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

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

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