Question

Evaluate the following Integral

Answer 1

split that into two integrals by writing the numerator as
\[ (x^2-1)+1\]

Answer 2

one will be a familiar integral and the other can be worked by a simple trig substitution

Answer 3

ahhh thank you

Answer 4

well, im still a little stuck

Answer 5

am i converting it to arcsin somehow?

Answer 6

yes the ifrst integral evaluates to arcsin

Answer 7

@ganeshie8 when you split it up, how do you write it out? I got confused :P

Answer 8

please, if we use this substitution:
\[x= \sin \theta \]
our integral can be write as below:
\[\int\limits \frac{ (\sin \theta )^{2} \cos \theta d \theta }{ (\cos \theta)^{3} }\]

Answer 9

\[\int\dfrac{x^2}{(1-x^2)^{3/2}}~dx = \int\dfrac{(x^2-1)+1}{(1-x^2)^{3/2}}~dx =\int\dfrac{-1}{(1-x^2)^{1/2}}~dx+\int\dfrac{1}{(1-x^2)^{3/2}}~dx \]

Answer 10

\[\int\dfrac{-1}{(1-x^2)^{1/2}}~dx+\int\dfrac{1}{(1-x^2)^{3/2}}~dx\] What happened to the \(x^2\)?

Answer 11

\[\dfrac{(x^2-1)+1}{(1-x^2)^{3/2}} = \dfrac{x^2-1}{(1-x^2)^{3/2}} + \dfrac{1}{(1-x^2)^{3/2}} \]

Answer 12

-1*

Answer 13

\[ = \dfrac{-(1-x^2)^1}{(1-x^2)^{3/2}} + \dfrac{1}{(1-x^2)^{3/2}} \]

right! exponent properties..

Answer 14

Yeah, gotcha :)

Answer 15

Yeah Iam slow, haha.

Answer 16

and subsequently we can write:
\[\int\limits \frac{ (\sin \theta )^{2} d \theta }{ (\cos \theta )^{2}}=\]
\[=\int\limits \left( \frac{ 1 }{ (\cos \theta )^{2} }-1 \right) d \theta \]

Answer 17

michele i dont understand how you got (1/cos^2 u) - 1

Answer 18

Yeah, I see how \(x=\sin(u)\) since the denominator can be written as \[\sqrt{({1-x^2})^3}\]And that can be seen as a trig identity.

Answer 19

Sorry it's taking a little while, wait a minute more please!

Answer 20

@Jgeurts
please substitute this, at numerator:
\[(\sin \theta)^{2}=1-(\cos \theta )^{2}\]

Answer 21

\[\int \frac{x^2}{(1-x^2)^{3/2}}dx\]Rewrite this , as ganeshie said, as :\[\int \frac{(x^2-1)+1}{(1-x^2)^{3/2}}dx\]Separate the fractions \[\int -\frac{\color{red}{(1-x^2)}}{\color{red}{(1-x^2)}\cdot (1-x^2)^{1/2}}dx+\int\frac{1}{(1-x^2)^{3/2}}dx\]\[\implies \int \frac{-1}{(1-x^2)^{1/2}}dx + \int \frac{1}{(1-x^2)^{3/2}} dx\]

Answer 22

Let me know if you understand what's going on here, then we can continue :)

Answer 23

im getting it now, im starting to like the trig way

Answer 24

but im understanding both

Answer 25

@Michele_Laino 's method? Sure, if that works, let him/her guide you to solving this integral :)

Answer 26

and i see now you have the integral of arcsin

Answer 27

actually it would be arccos because its negative right, at least your first part

Answer 28

No it would be \(\sin^{-1}(x)\).
\[\int \frac{du}{\sqrt{a^2-u^2}} =\sin^{-1}\left(\frac{u}{a}\right)+C\]

Answer 29

or a simplified version of that would be; \[\frac{1}{\sqrt{1-x^2}} = \sin^{-1}(x)\]

Answer 30

how are you solving the second portion?

Answer 31

please note that:
\[\int\limits \left( \frac{ 1 }{ (\cos \theta)^{2} } -1\right) d \theta=\]
\[\int\limits \frac{ d \theta }{ (\cos \theta )^{2} }-\int\limits d \theta =\]
\[= \tan \theta -\theta =\frac{ x }{ \sqrt{1-x ^{2}} }-\arcsin x\]

Answer 32

michele, genius!

Answer 33

@Jgeurts thank you!

Answer 34

\[\implies \int \frac{-1}{(1-x^2)^{1/2}}dx + \int \frac{1}{(1-x^2)^{3/2}} dx\]\[-\sin^{-1}(x) +\int \frac{1}{(1-x^2)^{3/2}} dx \]Let \(x=\sin(u)~,~ dx = \cos(u)du\)\[-\sin^{-1}(x)+\int \frac{1}{(\cos^2(u))^{3/2}}\cdot \cos(u)du\]\[-\sin^{-1}(x)+\int\frac{1}{\cos^2(u)}du\]\[-\sin^{-1}(x) + \int \sec^2(u)du\]\[=-\sin^{-1}(x) + \tan(x)+C\]

Answer 35

thanks everyone :)

Answer 36

@Jgeurts thanks!

Answer 37

Thank you :) Needed the practice.