Question

\(\color{green}{\textsf {Integrate :}}\)

\[\large \color{blue}{\int\limits \frac{x^2}{(x^2 + (z-c)^2)^\frac{3}{2}} dx}\]

Answer 1

z and c are constants ?

Answer 2

What substitution or rearrangement I should make here???

Yes, z and c are constants...

Answer 3

Looks like a trig substitution, pretend (z-c) is just a single thing.

Answer 4

That is surely not a problem, we can take (z-c) as some constant k also..

Answer 5

I agree @waterineyes but I don't know what's giving you problems or not.

The key to knowing what to choose for your trig substitution is just to look to see if anything is of this form: \[\Large \sin^2 \theta +\cos^2 \theta = 1\] See we can rearrange this to go from two things to just a single square to remove square roots simpler: \[\Large \cos^2 \theta = 1-\sin^2 \theta \]

Answer 6

adjust the numerator

x^2 = (x^2 +k^2) - k^2

then separate into 2 terms

Answer 7

see if that approach is shorter than any substitution....?

Answer 8

you can use the direct formula for 1/sqrt (x^2+a^2)
right ?

Answer 9

I try using that rearrangement....

Answer 10

For one term I can use, but I think next term is again creating problem for me..

Answer 11

I want to ask if the integral would be :

\[\int\limits \frac{dx}{(x^2 + k^2)^{\frac{3}{2}}}\]

Answer 12

I have a feeling that will be rough.

Let's just divide the pythagorean identity by cos^2 theta to get: \[\Large \frac{\sin^2 \theta}{\cos^2 \theta}+\frac{\cos^2 \theta}{\cos^2 \theta}=\frac{1}{\cos^2 \theta}\] this gives us:

\[\Large \tan^2 \theta + 1 = \sec^2 \theta\] which can be rearranged to give: \[\Large \tan^2 \theta = \sec^2 \theta - 1\]

Ok so now combine these last two with the original pythagorean identity and we essentially have a 1 to 1 correspondence between what we should substitute.

\[\Large y^2=1-x^2 \\ \Large y^2= 1+x^2 \\ \Large y^2 =-1+x^2\]

Answer 13

I think the next integral would be that I have shown above.. Yes??? @hartnn

Answer 14

So for example if we have: \[\Large \int\limits \frac{dx}{\sqrt{a^2-x^2}}dx\] \[\Large x=a \sin \theta\]

Answer 15

correct!
\(\int\limits \frac{dx}{(x^2 + k^2)^{\frac{3}{2}}}\)

a u-substitute could work here....if you feel this is getting complicated, go for trig

Answer 16

@Kainui, try to solve asked integral using trig substitution..

I am unable to think of any substitution here..

Answer 17

@hartnn yes integral is getting more complex and calculating if I do that, but I try..

Please you think of any other simpler one to reach there..

Answer 18

I have solved it, but you haven't really responded to me trying to help you out. Read what I wrote and I'll help you understand it, but I'm not going to give you the answer. \[\Large \int\limits \frac{x^2}{(x^2+(z-c)^2)^{3/2}}dx\]

Answer 19

@Kainui Hey don't give me the answer, actually I have one.. But I want to know how to reach there...

Answer 20

\(\int\limits \frac{dx}{(x^2 + k^2)^{\frac{3}{2}}}\)
is itself easy,
but yes, we finally have to do atrig substitution for this
i think x = a tan y will work

Answer 21

The example you are giving, there trig substitution is easy to be applied..

I want to know any subs that will work with asked integral..

Answer 22

x= k tan y
i mean

Answer 23

dx = ksec^2(y).dy

Answer 24

correct, denominatro, you will have sec^3

Answer 25

\[\frac{k \sec^2(y).dy}{\sec^3(y)} = \frac{\cos(y)}{k^2}\]

Answer 26

Yep, in second integral we can use substitution which is working successfully... :)

Answer 27

Now I solve it in neat and clean way... :)

Thanks to you @hartnn and @Kainui

Answer 28

welcome ^_^