Asked by Sean
How would I integrate
(4dt)/(sqrt(15-2t-t^2))
That thing has me stumped.
(4dt)/(sqrt(15-2t-t^2))
That thing has me stumped.
Answers
Answered by
drwls
Look for substitution x = bt + c than makes
(15-2t-t^2) become a^2 - x^2
The integral is an arcsin term.
-4 sin^-1[(30x-2)/56]
(15-2t-t^2) become a^2 - x^2
The integral is an arcsin term.
-4 sin^-1[(30x-2)/56]
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