Question

evaluate the integral
∫1/[16-x2(squared) )5/2].dx

Answer 1

\[\int\frac{1}{16-2x^2}\] ? Confused about that 5/2 on the side there...

Answer 2

( )5/2 that is squared of the bracket

Answer 3

oooh like \[\int\frac{1}{{(16-2x)}^{5/2}}\] ?

Answer 4

Evaluate the integral \[\int\frac{1}{{(16-2x)}^{5/2}}dx\]

Answer 5

Make it simpler. Turn it into:
\[\int(16-2x)^{-\frac{5}{2}}dx\] which is the same thing.

Answer 6

this is the sum\[\int\limits1/(16-x^2)^{5/2}\].dx

Answer 7

thanks for ur reply

Answer 8

can u solve this sum?

Answer 9

it's difficult to find the answer

Answer 10

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.

Answer 11

oops I forgot the dx in the integral..just pretend its there :P

Answer 12

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)

Answer 13

it's difficult to type the sums here

Answer 14

\[\int\frac{1}{\sqrt{(16-x^2)^5}}dx\]
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.

Answer 15

\[(16-x^2)^{5/2}=\sqrt{(16-x^2)^5}\]

Answer 16

ooh that's right

Answer 17

but,when we solve this,
\[\sqrt{16}(1-\sin ^{2}\theta)^{5}\] what is the answer? (square root of all)

Answer 18

it can't be 4cos theta because root 5 is there outside the 1- sin squared theta bracket

Answer 19

Try looking through these:
http://www.analyzemath.com/calculus/Differentiation/inverse_trigonometric.html

Answer 20

One of them should look close...

Answer 21

difficult to find.

Answer 22

try \[x=\frac{1}{4} sin x\] and \[dx=\frac{1}{4} cos dx\]

Answer 23

x=4sin theta
dx=4costheta.dtheta

Answer 24

\[x=4\sin \theta\]
\[dx=4\cos \theta.d \theta\]

Answer 25

i can solve up to that ..i need how to solve the final part.at last i got\[1/4 \int\limits \cos ^{-4} \theta.d \theta\]

Answer 26

then after that????

Answer 27

Hmm well actually if you make x=4 sin x and dx=4 cos dx it should work out ok because then you get:
\[\int\frac{4 cos \theta d\theta}{16^{5/2}(1-sin^2\theta)^{5/2}}\] =
\[\int\frac{cos \theta d\theta}{256(cos^2\theta)^{5/2}}\]

Answer 28

=\[\int\frac{cos \theta d\theta}{256(cos^5 \theta)}\] = \[\int\frac{d \theta}{256(cos^4 \theta)}\]

Answer 29

\[\frac{1}{256}\int{sec^4\theta} d \theta\]

Answer 30

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

Answer 31

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 \[sec^2\theta=1+tan^2\theta\]