Integrating 8sin4x: A Step-by-Step Solution Guide

[jdawg]

Homework Statement

∫8sin⁴x dx

Homework Equations

The Attempt at a Solution

8∫sin²xsin²x

I'm not really sure what to do next. Maybe substitute a 1-cos²x in for one of the sin²x? Maybe both?

 

[Dick]

Homework Statement

∫8sin⁴x dx

Homework Equations

The Attempt at a Solution

8∫sin²xsin²x

I'm not really sure what to do next. Maybe substitute a 1-cos²x in for one of the sin²x? Maybe both?

That sort of integration involves using double angle identities in trig. Look them up and try and get started.

 

[ehild]

Homework Statement

∫8sin⁴x dx

Homework Equations

The Attempt at a Solution

8∫sin²xsin²x

I'm not really sure what to do next. Maybe substitute a 1-cos²x in for one of the sin²x? Maybe both?

Do you know the relations

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

ehild

 

[jdawg]

Ok, so I plugged in (1-cos(2x))/2 for both of my sin²x and then foiled.
This is what I have now:
2∫1+cos²2x+2cos2x dx

 

[Dick]

Ok, so I plugged in (1-cos(2x))/2 for both of my sin²x and then foiled.
This is what I have now:
2∫1+cos²2x+2cos2x dx

No, that's not what you get. There's a sign problem. But the important point is that you can use the double angle identities on $\cos^2(2x)$ as well.

 

[jdawg]

Oh, so it should be: 2∫1+cos²2x-2cos2x dx
Then: 2∫cos²2x-cos2x dx
Now could you split up your integrals and solve them individually?

 

[Dick]

Oh, so it should be: 2∫1+cos²2x-2cos2x dx
Then: 2∫cos²2x-cos2x dx
Now could you split up your integrals and solve them individually?

Of course you can split them up and solve them individually. That's the whole point. But now what happened to the "1" part? And a factor of 2 also disappeared. Just take it step by step.