How to integrate a trig function of even powers?

The real question is how do you solve this integral? What help do you need exactly? Where are you stuck? Try something, show it to us and we will guide you towards the solution.
How do you think that you can get an odd power of 6? What is half of 6? Maybe use the half angle theorem.

Please post back with your work.
 
Yes, I initially tried IBP, but wound up with the original integral being equal to the original integral. Half-angle identities led to success. :)
 
Let's follow up here:

Let:

[MATH]I=\int \cos^4(x)\sin^6(x)\,dx[/MATH]
Consider the following two identities:

[MATH]\sin^2(x)=\frac{1-\cos(2x)}{2}[/MATH]
[MATH]\cos^2(x)=\frac{1+\cos(2x)}{2}[/MATH]
These identities allow us to write the integrand as:

[MATH]\frac{1}{32}(1-\cos(2x))^3(1+\cos(2x))^2=\frac{1}{32}(1-\cos(2x))(1-\cos^2(2x))^2=\frac{1}{32}(1-\cos(2x))\sin^4(2x)[/MATH]
Applying the first of the two identities, we obtain:

[MATH]\frac{1}{128}(1-\cos(2x))(1-\cos(4x))^2=\frac{1}{128}(1-\cos(2x))(1-2\cos(4x)+\cos^2(4x))[/MATH]
Applying the second of the identities above, we get:

[MATH]\frac{1}{256}(1-\cos(2x))(3-4\cos(4x)+\cos(8x))[/MATH]
Expanding, we obtain:

[MATH]\frac{1}{256}(3-4\cos(4x)+\cos(8x)-3\cos(2x)+4\cos(2x)\cos(4x)-\cos(2x)\cos(8x))[/MATH]
Next, consider the following product-to-sum identity:

[MATH]2\cos(\alpha)\cos(\beta)=\cos(\alpha-\beta)+\cos(\alpha+\beta)[/MATH]
Applying this, our integrand becomes:

[MATH]\frac{1}{512}(6-2\cos(2x)-8\cos(4x)+3\cos(6x)+2\cos(8x)-\cos(10x))[/MATH]
Now, we have an integrand that can be directly integrated:

[MATH]I=\frac{1}{512}\left(6x-\sin(2x)-2\sin(4x)+\frac{1}{2}\sin(6x)+\frac{1}{4}\sin(8x)-\frac{1}{10}\sin(10x)\right)+C[/MATH]
And finally:

[MATH]I=\frac{1}{10240}\left(120x-20\sin(2x)-40\sin(4x)+10\sin(6x)+5\sin(8x)-2\sin(10x)\right)+C[/MATH]
 
I attempted the problem again, and these are my steps. I did not get the same answer as MarkFL. I don't know where I went wrong.
These are my steps - I used half angle identities:
[MATH]I=\int\cos^4x\sin^6x\ dx[/MATH]using the identities: [MATH]\sin x\cos x=\frac{\sin(2x)}{2}[/MATH] and [MATH]\sin^2x=\frac{1-\cos(2x)}{2}[/MATH]:
[MATH]\int\left(\frac{\sin(2x)}{2}\right)^4\left(\frac{1-\cos(2x)}{2}\right)\ dx[/MATH][MATH]\frac{1}{32}\int(\sin^4(2x))(1-\cos(2x))\ dx = \frac{1}{32}\int\sin^4(2x)-(\cos(2x))(\sin^4(2x))\ dx[/MATH]We can now divide this integral into two parts: [MATH]\frac{1}{32}\int\sin^4(2x)\ dx [/MATH] and [MATH]\frac{-1}{32}\int\cos(2x)\sin^4(2x)\ dx[/MATH]Let's integrate the first part first: [MATH]\frac{1}{32}\int\sin^4(2x)\ dx [/MATH][MATH]=\frac{1}{32}\int(1-\cos^2(2x))(\sin^2(2x)\ dx[/MATH][MATH]=\frac{1}{32}\int(\sin^2(2x)-\sin^2(2x)(\cos^2(2x))\ dx[/MATH][MATH]=\frac{1}{32}\int\frac{1-\cos(4x)}{2}-\frac{\sin^2(4x)}{4}\ dx[/MATH][MATH]=\frac{1}{128}\int2-2\cos(4x)-\sin^2(4x)\ dx[/MATH][MATH]=\frac{1}{128}\int2-2\cos(4x)-\frac{1-\cos(8x)}{2}\ dx[/MATH][MATH]=\frac{x}{64}-\frac{\sin(4x)}{256}-\frac{1}{256}\int1-\cos(8x)\ dx[/MATH][MATH]=\frac{x}{64}-\frac{\sin(4x)}{256}-\frac{x}{256}+\frac{\sin(8x)}{2048}[/MATH]Now, the second part: [MATH]\frac{-1}{32}\int\cos(2x)\sin^4(2x)\ dx[/MATH]Let's do a u-substitution [MATH]u=\sin(2x)\ \ \ \ \ \ \ \ \frac{du}{dx}=2\cos(2x)\ \ \ \ \ \ \ \ \frac12du=\cos(2x)dx[/MATH]Now we have: [MATH]\frac{-1}{64}\int u^4\ du[/MATH][MATH]\frac{-u^5}{320} = \frac{-\sin^5(2x)}{320}[/MATH] ........................Edited
Now, putting everything together:
[MATH]I=\frac{x}{64}-\frac{\sin(4x)}{256}-\frac{x}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{320}+C[/MATH]........................Edited

[MATH]=\frac{3x-\sin(4x)}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{320}+C[/MATH]........................Edited
 
Last edited by a moderator:
It looks like you began with a power of 2 on the sine function instead of 6 as your original post indicates.
 
I don't think I did - 4 of the sines are in the first term which is to the fourth power. The other two are in the second term. The mistake was only in writing out the problem on the top.
 
Look at the last line of the part in red, where you back substitute...you have u to the 5th power, but the sine function is only to the 1st power.
 
No, it's not:

[MATH]\int\cos(2x)\sin^4(2x) dx[/MATH] is what's being integrated. u=sin(x) to the first power, that's true.
 
burt said:
[MATH]\frac{-u^5}{620} = \frac{-\sin(2x)}{620}[/MATH]
Click to expand...
You missed the 5th power for the sin function
Remember two things. MarkFL's result can be equivalent to your result but still look different and they can differ by a constant as well.
 
whoops - you are right.
Other than that, is my work correct?
 
burt said:
whoops - you are right.
Other than that, is my work correct?
Click to expand...
There was quite a bit of work but I only noticed that one mistake. You can always take the derivative of your answer and see if you get back the integrand.
 
This was the calculators answer for the derivative. It does not seem to be the same thing...
[MATH]\dfrac{155\cos\left(8x\right)-620\cos\left(4x\right)-512\cos\left(2x\right)\sin^3\left(2x\right)+465}{39680}[/MATH]
 
Correction on the second part (red) part: as @Jomo said, I forgot to raise the sine to the 5th power. But, the denominator is also incorrect. It should be:
[MATH]\frac{-\sin^5(2x)}{320}[/MATH]
 
Okay when I use:

[MATH]I=\frac{3x-\sin(4x)}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{320}+C[/MATH]
W|A says that this is equivalent to the result I posted. :)
 
MarkFL said:
Okay when I use:

[MATH]I=\frac{3x-\sin(4x)}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{320}+C[/MATH]
W|A says that this is equivalent to the result I posted. :)
Click to expand...
That takes care of things, then :D. Question though - what is W|A? Wolfram Alpha?
 
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