Question

Integral sin x upon sin 4x DX

Answer 1

I think the approach to solve this problem is first to expand the denominator that is

Sin4x=2sin2xcos2x. Then write sin2x=2sinxcosx then the denominator will be 4sinxcosxcos2x.

Now cancel sinx from numerator and denominator.

You are left with 1/cosxcos2x

Now this part is the trickiest

Mutipl y cosx in the numerator and denominator.

You get cosx/cos^2x*cos2x

Replace cos^2x=1-sin^2x

Cos2x =1–2sin^x

Then replace sinx=t

Then. Cosxdx=DT

So now this changes to DT/(1-t^2)*(1–2t^2 )

Then I think you can use partial fraction to solve this