13:: Is it not an hypergeometric Function ! Substituting x=2sina gives u
\dpi{120} I=\sqrt[3]{2sin3a}. \, \, 2sina \, da
5 Answers
62First one: \frac{x^3-x}{(x^2+1)^3+4(x^4+x^2)}=\frac{1-1/x^2}{(x+1/x)^3+4(x+1/x)}
623) the standard trick that we often use to convert this to cos 3 theta
x=kt
3x-x3=3kt - k3t3=k(3t - k2t3)
so for the substitution of sin theta, we want k2=4
so we substitute k=2
so we will end up with x=2t
\int^._. ^3\sqrt{2(3t-4t^3)}.2dt
Substitute t= sin theta, thsi will get converted to sin 3 theta
here this method will not give as good benefits as i initally thought.. but now that i have typed so much the idea can be tehre on how to convert a cubic to cos 3x form :P
1
262for 2nd problem, just multiply numberator and denominator with x.
bring x inside the root in the denominator.
then take the part inside the denominator as t.
then dt will be a constant mutiple of the numerator.
\int \frac{x^{2012}+x^{2011}}{\sqrt[2012]{2012.x^{2013}+2013.x^{2012}}}dx
now 2012.x^{2013}+2013.x^{2012} = t
2012.2013.(x^{2011}+x^{2012})dx = dt
nothing more to say after that.
