anonymous
  • anonymous
evaluate the integral ∫1/[16-x2(squared) )5/2].dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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raspberryjam
  • raspberryjam
\[\int\frac{1}{16-2x^2}\] ? Confused about that 5/2 on the side there...
anonymous
  • anonymous
( )5/2 that is squared of the bracket
raspberryjam
  • raspberryjam
oooh like \[\int\frac{1}{{(16-2x)}^{5/2}}\] ?

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2bornot2b
  • 2bornot2b
Evaluate the integral \[\int\frac{1}{{(16-2x)}^{5/2}}dx\]
raspberryjam
  • raspberryjam
Make it simpler. Turn it into: \[\int(16-2x)^{-\frac{5}{2}}dx\] which is the same thing.
anonymous
  • anonymous
this is the sum\[\int\limits1/(16-x^2)^{5/2}\].dx
anonymous
  • anonymous
thanks for ur reply
anonymous
  • anonymous
can u solve this sum?
anonymous
  • anonymous
it's difficult to find the answer
raspberryjam
  • raspberryjam
Argh it's a trigonometry integral...I really hate these types of problems too. \[\int\frac{1}{\sqrt{(16-x^2)^5}}\] is what you've got. This should look familiar to you...look at the derivatives of arcsin, arctan, arccos, etc.
raspberryjam
  • raspberryjam
oops I forgot the dx in the integral..just pretend its there :P
anonymous
  • anonymous
i too hate those problems.the sum is like this.. \[\int\limits1/(16-x ^{2})\left(\begin{matrix}5/2 \\ ?\end{matrix}\right)\].dx (no question mark below 5/2)
anonymous
  • anonymous
it's difficult to type the sums here
raspberryjam
  • raspberryjam
\[\int\frac{1}{\sqrt{(16-x^2)^5}}dx\] This is the same thing as what you have written. Look at the derivatives of arcsin, arctan, arccos, etc. One of them should look similar.
raspberryjam
  • raspberryjam
\[(16-x^2)^{5/2}=\sqrt{(16-x^2)^5}\]
anonymous
  • anonymous
ooh that's right
anonymous
  • anonymous
but,when we solve this, \[\sqrt{16}(1-\sin ^{2}\theta)^{5}\] what is the answer? (square root of all)
anonymous
  • anonymous
it can't be 4cos theta because root 5 is there outside the 1- sin squared theta bracket
raspberryjam
  • raspberryjam
Try looking through these: http://www.analyzemath.com/calculus/Differentiation/inverse_trigonometric.html
raspberryjam
  • raspberryjam
One of them should look close...
anonymous
  • anonymous
difficult to find.
raspberryjam
  • raspberryjam
try \[x=\frac{1}{4} sin x\] and \[dx=\frac{1}{4} cos dx\]
anonymous
  • anonymous
x=4sin theta dx=4costheta.dtheta
anonymous
  • anonymous
\[x=4\sin \theta\] \[dx=4\cos \theta.d \theta\]
anonymous
  • anonymous
i can solve up to that ..i need how to solve the final part.at last i got\[1/4 \int\limits \cos ^{-4} \theta.d \theta\]
anonymous
  • anonymous
then after that????
raspberryjam
  • raspberryjam
Hmm well actually if you make x=4 sin x and dx=4 cos dx it should work out ok because then you get: \[\int\frac{4 cos \theta d\theta}{16^{5/2}(1-sin^2\theta)^{5/2}}\] = \[\int\frac{cos \theta d\theta}{256(cos^2\theta)^{5/2}}\]
raspberryjam
  • raspberryjam
=\[\int\frac{cos \theta d\theta}{256(cos^5 \theta)}\] = \[\int\frac{d \theta}{256(cos^4 \theta)}\]
raspberryjam
  • raspberryjam
\[\frac{1}{256}\int{sec^4\theta} d \theta\]
raspberryjam
  • raspberryjam
Uhh oh man.. \[\frac{1}{256}\int{(1+tan^2\theta)sec^2\theta} d \theta\] = \[\frac{1}{256}\int{sec^2\theta+tan^2\theta sec^2\theta} d \theta\] then you have to split it =\[\frac{1}{256}\int{(tan^2\theta sec^2\theta) d\theta}+\int{sec^2\theta} d \theta\] use \[\theta=tan\theta\] and \[d\theta=sec^2\theta d\theta\] I think you should be able to get it from here.. It helps if you have this in front of you by the way: http://www.sosmath.com/trig/Trig5/trig5/trig5.html
raspberryjam
  • raspberryjam
kind of skipped a step but you can see that \[\frac{1}{256}\int{sec^4\theta} d \theta\] = \[\frac{1}{256}\int{(sec^2\theta)(sec^2\theta)} d \theta\] and based on a trig identity \[sec^2\theta=1+tan^2\theta\]

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