SOLUTION: What is the integral of (x^2)(√5-x^2)dx using trigonometric substitution?
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Question 1193638: What is the integral of (x^2)(√5-x^2)dx using trigonometric substitution?
Answer by Solver92311(821) (Show Source): You can put this solution on YOUR website!
\,dx)
looks like 
, so substitute )
where 
and 
\)\ =\ \sqrt{5}\frac{d}{du}\sin(u)\ =\ \sqrt{5}\cos(u))
\,dx\ =\ \int\,\(\sqrt{5}\sin(u)\)^2\sqrt{5\,-\,\(\[\sqrt{5}\sin(u)\]^2\)}\,\(\sqrt{5}\cos(u)\)\,du)
Simplify:
\)\,\sqrt{5\,-\,5\sin^2(u)}\,\sqrt{5}\cos(u)\,du)
Note: }\ =\ \cos(\varphi))
\)\,\sqrt{5}\cos(u)\,\sqrt{5}\cos(u)\,du\ =\ \int\,25\sin^2(u)\cos^2(u)\,du)
Note: }{8})
\)\,du)
\,du\])
\]\ +\ C)
)
\ -\ \frac{1}{4}\,\sin\(4\arcsin\(\frac{1}{\sqrt{5}}x\)\)\]\ +\ C)
Or if you prefer
\ -\ \frac{1}{4}\,\sin\(4\sin^{-1}\(\frac{1}{\sqrt{5}}x\)\)\]\ +\ C)
John
My calculator said it, I believe it, that settles it
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