some integrals

5 Answers

62
Lokesh Verma ·

First one: (x2+1)3+4(x4+x2)x3x=(x+1/x)3+4(x+1/x)11/x2

62
Lokesh Verma ·

3) 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
narayan ·

3:: Is it not an hypergeometric Function ! Substituting x=2sina gives u

\dpi{120} I=\sqrt[3]{2sin3a}. \, \, 2sina \, da

262
Aditya Bhutra ·

for 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.

[2012]2012.x2013+2013.x2012x2012+x2011dx

now 2012.x2013+2013.x2012=t

2012.2013.(x2011+x2012)dx=dt

nothing more to say after that.

1708
man111 singh ·

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