Yes there is the standard substitution for these kinds of integrals. And the substitution is tan (@/2)= t. Differentiating that u get d@=2/(1+t^2)dt. Because differential of tan(@/2) is 1/2*1/cos(@/2)^2 and this is equal to1*2( 1+Tan(@/2)^2).This is the usual identity just refresh you knowledge...
Hey there,
The trick is to use identity 1+Z^3=(1+z)(1-Z+Z^2) and then just square this expression and do the partial fraction as usual.
Hope this helps:)))
-Suppose you want to integrate something like arcsinx* x/Sqrt(1-x^2)dx
1) recognise that x/Sqrt(1-x^2) is in fact derivative of -Sqrt(1-x^2)
2) Then You have in the integral -arcsinx*d(Sqrt(1-x^2)dx
3)Using partial integration you then have
-arcsinx*Sqrt(1-x^2)+integral...
Yes I just realized that you can get nicer expression if you see that
cos2x=1/Sqrt(1+tan^2(2x)) then cos2x=1-2sin^2(x)
hence
1-2sin^2(x)=1/(Sqrt(1+tan^2(2x)))
Well tan2x=Sin2x/Cos2x then
Sin2x= Tan2x*Cos2x but note that Cos^2(2x)=1/Sec^2(2x) using sec^2(2x)=1+ tan^2(2x) we then get
Sin2x=Tan2x/Sqrt(1+tan^2(2x)) this is all ok but sin2x=2sinxcosx so you need to do the same for cos2x and find cosx in terms of tan2x thus replace it into the expression...