Question

Trig sub please help

Answer 1

\[\int\limits \frac{ 1 }{ x ^{2} \sqrt{x ^{2} +1}} dx\]

Answer 2

Do you know which trig sub to use?

Answer 3

\[\Large\rm \color{orangered}{x^2}+1\]
Which one seems more appropriate for this problem?\[\Large\rm 1-\color{orangered}{\sin^2\theta}=\cos^2\theta\]\[\Large\rm \color{orangered}{\tan^2\theta}+1=\sec^2\theta\]

Answer 4

we use tan theta?

Answer 5

Mmmm k good good good.

Answer 6

\[\Large\rm x=\tan \theta, \qquad\qquad dx=?\]

Answer 7

\[\Large\rm \int\limits\limits \frac{ 1 }{ \color{orangered}{x}^{2} \sqrt{\color{orangered}{x}^{2} +1}} dx=\int\limits\limits \frac{ 1 }{ \color{orangered}{\tan}^{2}\theta \sqrt{\color{orangered}{\tan}^{2}\theta +1}} dx\]

Answer 8

\[\sec ^{2}\theta dx\]

Answer 9

\(\Large\sec ^{2}\theta~d\theta\)*
Ok good.

Answer 10

\[\Large\rm \int\limits\limits\limits \frac{ dx }{ \color{orangered}{\tan}^{2}\theta \sqrt{\color{orangered}{\tan}^{2}\theta +1}}=\int\limits\limits\limits \frac{ \sec^2\theta~d \theta }{ \color{orangered}{\tan}^{2}\theta \sqrt{\color{orangered}{\tan}^{2}\theta +1}}\]

Answer 11

So what's happening under the root?

Answer 12

were plugging in tan theta for x and sec^2 theta for dx right

Answer 13

Yes.
I moved the dx to the top of the fraction so it would be easier to fit on the screen >.<

Answer 14

so now I have to find the integral of tan^2θ ?

Answer 15

Under the root we want to apply our Pythagorean Identity,\[\Large\rm \tan^2\theta+1=\sec^2\theta\]
\[\Large\rm \int\limits\frac{ \sec^2\theta~d \theta }{ \tan^{2}\theta \sqrt{\tan^{2}\theta +1}}=\int\limits\frac{ \sec^2\theta~d \theta }{ \tan^{2}\theta \sqrt{\sec^2\theta}}\]
Cancel out a secant from top and bottom:\[\Large\rm\int\limits\limits\frac{ \sec\theta~d \theta }{ \tan^{2}\theta}\]Integral of tan^2theta? Hmm I'mm not sure where you're getting that from.

Answer 16

sorry what i meant was changing \[\tan ^{2}\theta \to \sec ^{2}\theta \]

Answer 17

\(\large \tan ^{2}\theta+1 \to \sec ^{2}\theta\)
you mean? >.<

Answer 18

So if you were able to follow those steps, looks like we end up with:\[\Large\rm\int\limits\limits\limits\frac{ \sec\theta~d \theta }{ \tan^{2}\theta}\]And then from there, ummmm...

Answer 19

I guess writing it in terms of sines and cosines will help get through this part.

Answer 20

yes

umm im trying to solve now

Answer 21

having trouble

Answer 22

Converting to sines and cosines:\[\Large\rm\int\limits\frac{\cos^2\theta}{\cos \theta \sin^2\theta}d \theta\]Understand what I did there?

Answer 23

secant is one over cosine, so I threw it into the bottom.
tangent is sine over cosine, but since it's in the denominator it gets flipped.

Answer 24

Canceling some stuff,\[\Large\rm \int\limits \frac{\cos \theta}{\sin^2\theta}d \theta\]

Answer 25

u-substitution: \(\Large\rm \quad u=\sin\theta, \qquad\qquad du=\cos\theta~d\theta\)

Answer 26

Where are you getting stuck at?

Answer 27

i was having trouble substituting by trig sub, which i need to brush up on so now

not to familiar with trig sub yet just been introduced to the topic so having trouble on going through the steps

Answer 28

I gotta get going :c

But maybe you want to try starting with something a little easier for now?
Like:\[\Large\rm \int\limits \frac{dx}{x^2+1}\]Same substitution, but less things to plug in. And it simplifies down a lot nicer.
I just can't seem to pin point where you're getting stuck.

Ok sorry bye bye c: I'll be back later if you still need help.