Question

Integral Calculus ....

Find the value of integral :

\( I = \int \left( \cfrac{x^2}{1+x^6} dx \right) \)

Answer 1

@ganeshie8 @BSwan @UnkleRhaukus

Answer 2

x^3=tan(theta)

Answer 3

I don't get it that why do we do this? Is there any specific rule for this?

Answer 4

http://i.imgur.com/11hGmBW.png

Answer 5

The green rectangle.

Answer 6

Do you have an integral?

Answer 7

My first attempt was just to set x^2=u, but I realized that wouldn't work.

Usually when I see something similar to \[1\ + x^2\] I try to see if I can fit it to a trig substitution. In this case, \[\large x^6=(x^3)^2\] so I just tried it out. It also seems to make sense that this is right because the derivative of x^3 is x^2.

Answer 8

^ derivative is 3x^2...

And okay, so, the next step will to replace dx by d(tan(theta)) ?

Answer 9

Well I didn't mean that literally, I just meant that x^3 has a scalar multiple derivative of x^2 but whatever lol

Yeah. Just take it through like a normal substitution now.

Answer 10

That comes out to be :

I = \(\int \left(\cfrac{1}{3(1+\tan^2 \theta )} d\tan \theta\right) \)

Answer 11

\(d(\tan \theta) =3x^2 dx\)

\(dx = \cfrac{d(\tan \theta)}{3x^2}\)

So, I just plugged this in at the place of dx in the integral and I got :

\(\color{blue}{\text{Originally Posted by}}\) @mathslover
That comes out to be :

I = \(\int \left(\cfrac{1}{3(1+\tan^2 \theta )} d\tan \theta\right) \)
\(\color{blue}{\text{End of Quote}}\)

Answer 12

\[\large x^3=\tan \theta \\ \large 3x^2dx = \sec^2 \theta d \theta\]
I'd write it more like this I think.

Answer 13

\( I = \cfrac{1}{3} \int (\cos^2 \theta )d\tan \theta \)

Answer 14

Hmm well yes, youright...

\(I = \cfrac{1}{3} (\cos^2 \theta) \sec^2 \theta d\theta\)

Right?

Answer 15

You got it. =)

Answer 16

Yes, but I missed integral sign there...

Answer 17

Yeah but you left the dTheta in there so I assumed you knew it wasn't over lol.

Answer 18

As the answer comes out to be \(\cfrac{\theta}{3}\)

So, finally :

\(\cfrac{1}{3} \times \left(\tan^{-1} (x^3) \right) + C\)

That should be the final answer, right?

Answer 19

Yep. It sorta worked out much nicer than expected haha.

Answer 20

Haha yeah... Thanks for the help.

Answer 21

Yeah, glad I could help. =)

Answer 22

let x^3=z

Answer 23

Yeah shamim, I just noticed that we can do it like that too :

Let \(x^3 = u\)

du = \(3x^2 dx\)

\(dx = \cfrac{du}{3x^2}\)

That becomes : \(I = \int \cfrac{x^2}{1+(x^3)^2} dx = \int \cfrac{x^2}{1 + u^2} \times \cfrac{du}{3x^2} \)

or : \(\int \cfrac{du}{(1+u^2)3} \)

Since : \(\int \cfrac{1}{1+x^2} \times dx\) = \(\tan^{-1} x + C\)

Therefore, I = \(\cfrac{1}{3} \tan^{-1} (x^3) + C \)

Answer 24

Though, @Kainui 's method was also nice. :-)

Answer 25

The inverse trig integrals are just a special case of trig substitution anyways. =P