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Help Trigonometric substitution \[\int\limits_{5}^{10}\sqrt{100-x ^{2}}dx\]
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same hint : a^2-x^2 put x= a sin theta
let x = 10sin(x)
\[\sqrt{100-(10\sin \theta)^{2}}dx\]
\[\sqrt{100-100\sin ^{2}x}\]
Then use a trig identity
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then dx= 10 cos t
Yes, let me work out the full problem real quick
yes, ans what does sqrt (....) simplify to ?
so will it eventually just be 100 cos^2 t
yes. and you know the cos 2x formula ?
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cos^2 t = (1+cos2 t)/2 that one?
yes
yes, that should make it easier to solve now :)
alright thanks so much again
ask if you get stuck again anywhere....
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