How do I find the anti derivative of (Sin(x))^4....?

Question

How do I find the anti derivative of (Sin(x))^4....?

Answer 1

When finding antiderivatives of even powers of sine, the trick is to keep this identity in mind:

[sin(x)]^2 = (1 - cos2x)/2

The right hand side is SO much easier to integrate than the left hand side. That's what we're going to do here.

Note that

[sin(x)]^4 = { [sin(x)]^2 }^2

So all we have to do is plug in our identity in those squiggly brackets.

[ (1 + cos2x)/2 ]^2

Which becomes

(1 + 2cos2x + [cos(2x)]^2) / 4

Now, we apply the same identity to (cos(2x))^2, only this time, it's a bit different

[cos(2x)]^2 = (1 + cos[2(2x)])/2

See what I did there? It's the SAME formula, except you replace each x by 2x. This results in

[cos(2x)]^2 = (1 + cos[4x)])/2

And now, we substitute what we just solved for in here:

(1 + 2cos2x + [cos(2x)]^2) / 4
(1 + 2cos2x + [1 + cos[4x]]/2 ) /4

Since that third term is divided by 2, AND since this is a complex fraction, we multiply the top and bottom by 2, to get
(2 + 4cos2x + [1 + cos[4x]]) / 8

And we can actually reduce this further by combining like terms.

(3 + 4cos2x + cos4x)/8

At this point, we didn't even START to integrate yet. And now it's about time we do. We went from here:

Integral ([sin(x)]^4)dx

To here:

Integral ((3 + 4cos2x + cos4x)/8)dx

Let's pull out the 1/8, because it's a constant. Thus, we get

(1/8) * Integral (3 + 4cos2x + cos4x)dx

Which we can split into three integrals.

(1/8) [Integral (3 dx) + 4*Integral (cos2x)dx + Integral (cos4x)dx ]

From here, the integral is easy to solve.

(1/8) [3x + 4[(sin2x)/2] + (sin(4x))/4] + C

[3x + 2sin2x + (sin4x)/4 ]/8 + C
[12x + 8sin2x + sin4x]/32 + C

Answer 2

put it as ((sinx)^2)^2 then (sinx)^2=(1-cos(x/2))/2
expand tht and integrate the acquired equation