I went for 3 substitutions,
\( tanx=u^{1/3} \),
\(u=tan\theta\),
\( \cos2\theta=t\). Some identities which i used are,
\[1+x^6=(1+x^2)(x^4-x^2+1) \] \[1+\cos2\theta=2\cos^{2}\theta \] \[1-\cos2\theta=2\sin^{2}\theta\] \[\sin2\theta=2\sin\theta\cos\theta\] \[tan^{-1}x + tan^{-1}y=tan^{-1}{\dfrac{x+y}{1-xy}} \]
I think this will get you through.
–
Rudraksh Shukla
·
8 months, 2 weeks ago

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TopNewestI went for 3 substitutions, \( tanx=u^{1/3} \), \(u=tan\theta\),

\( \cos2\theta=t\). Some identities which i used are, \[1+x^6=(1+x^2)(x^4-x^2+1) \] \[1+\cos2\theta=2\cos^{2}\theta \] \[1-\cos2\theta=2\sin^{2}\theta\] \[\sin2\theta=2\sin\theta\cos\theta\] \[tan^{-1}x + tan^{-1}y=tan^{-1}{\dfrac{x+y}{1-xy}} \] I think this will get you through. – Rudraksh Shukla · 8 months, 2 weeks ago

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