# Thread: Indefinite integral of cos^4 2x dx

1. ## Indefinite integral of cos^4 2x dx

I tried integrating this on my own but couldn't get the correct answer. I'm not supposed to use any reduction formulas. Any help is much appreciated

Integral (cos42x dx)
Integral (1/2(1-cos 4x))2
Integral ([1/4(1-2cos 4x- cos24x)][1/4(1-2cos 4x - cos 24x)]dx)
1/8 integral ((1-2cos4x - cos 24x)(1-2cos 4x - cos 24x) dx)
1/8 integral (1-2cos 4x - cos24x-2 cos 4x-4cos 24x - 2 cos 34x - cos24x - 2 cos34x - cos44x
1/8 integral (1-4cos4x-6cos24x - 4 cos34x -cos44x dx)

I stopped here because it wouldn't give me the answer I needed. The answer is
1/64 (24x + 8 sin 4x + 8 sin x) + C

2. Look at this website.
B
e sure to click the show steps button.

3. Hello, Ohoneo!

$\displaystyle \int \cos^42x\,dx$

We have: .$\cos^4\!2x \:=\:(\cos^2\!2x)^2 \:=\:\left(\frac{1+\cos4x}{2}\right)^2 \;=\;\frac{1}{4}(1 + 2\cos4x + \cos^2\!4x)$

. . . . . . $=\;\frac{1}{4}\left(1 + 2\cos4x + \frac{1+\cos8x}{2}\right) \;=\;\frac{1}{4}\left(\frac{3}{2} + 2\cos4x + \frac{1}{2}\cos8x\right)$

Then: .$\dfrac{1}{4}\displaystyle \int\left(\tfrac{3}{2} + 2\cos4x + \tfrac{1}{2}\cos8x\right)\,dx \;=\; \tfrac{1}{4}\left(\tfrac{3}{2}x + \tfrac{1}{2}\sin4x + \tfrac{1}{16}\cos8x\right) + C$

4. Originally Posted by soroban
Hello, Ohoneo!

We have: .$\cos^4\!2x \:=\\cos^2\!2x)^2 \:=\:\left(\frac{1+\cos4x}{2}\right)^2 \;=\;\frac{1}{4}(1 + 2\cos4x + \cos^2\!4x)$

. . . . . . $=\;\frac{1}{4}\left(1 + 2\cos4x + \frac{1+\cos8x}{2}\right) \;=\;\frac{1}{4}\left(\frac{3}{2} + 2\cos4x + \frac{1}{2}\cos8x\right)$

Then: .$\dfrac{1}{4}\displaystyle \int\left(\tfrac{3}{2} + 2\cos4x + \tfrac{1}{2}\cos8x\right)\,dx \;=\; \tfrac{1}{4}\left(\tfrac{3}{2}x + \tfrac{1}{2}\sin4x + \tfrac{1}{16}\cos8x\right) + C$
Hey I was just wondering where you got the (3/2) from after you foil both half angle identities?

5. Originally Posted by esahc
Hey I was just wondering where you got the (3/2) from after you foil both half angle identities?
Not to be rude answering for Soroban, but the 3/2 comes from the sum of 1 and 1/2 within the first term. Remember that $\frac{1+cos8x}{2}$ can be written as $\frac{1}{2}+\frac{cos8x}{2}$

6. Oh i see now thanks for the help

7. Originally Posted by Ohoneo
I tried integrating this on my own but couldn't get the correct answer. I'm not supposed to use any reduction formulas. Any help is much appreciated

Integral (cos42x dx)
Integral (1/2(1-cos 4x))2
Okay to here.
Integral ([1/4(1-2cos 4x- cos24x)][1/4(1-2cos 4x - cos 24x)]dx)
?? (1/4(1- 2cos(4x)- cos2(4x)) is (1/2(1- cos(4x))^2. Why do you have it twice?

1/8 integral ((1-2cos4x - cos 24x)(1-2cos 4x - cos 24x) dx)
1/8 integral (1-2cos 4x - cos24x-2 cos 4x-4cos 24x - 2 cos 34x - cos24x - 2 cos34x - cos44x
1/8 integral (1-4cos4x-6cos24x - 4 cos34x -cos44x dx)

I stopped here because it wouldn't give me the answer I needed. The answer is
1/64 (24x + 8 sin 4x + 8 sin x) + C

8. ## In the last expression of the final integral, 'cos 8x' must be replaced by 'six 8x'.

Originally Posted by soroban
Hello, Ohoneo!

We have: .$\cos^4\!2x \:=\\cos^2\!2x)^2 \:=\:\left(\frac{1+\cos4x}{2}\right)^2 \;=\;\frac{1}{4}(1 + 2\cos4x + \cos^2\!4x)$

. . . . . . $=\;\frac{1}{4}\left(1 + 2\cos4x + \frac{1+\cos8x}{2}\right) \;=\;\frac{1}{4}\left(\frac{3}{2} + 2\cos4x + \frac{1}{2}\cos8x\right)$

Then: .$\dfrac{1}{4}\displaystyle \int\left(\tfrac{3}{2} + 2\cos4x + \tfrac{1}{2}\cos8x\right)\,dx \;=\; \tfrac{1}{4}\left(\tfrac{3}{2}x + \tfrac{1}{2}\sin4x + \tfrac{1}{16}\sin8x\right) + C$

In the last expression of the final integral, 'cos 8x' must be replaced by 'six 8x'.

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