Question

Trig Substitution (Calc)

Answer 1

\[\int\limits_{}^{} \frac{ \sqrt{16+x^2} }{ x^2 }dx\]

Answer 2

So this is what I have for the next step:
\[\int\limits_{}^{}\frac{ 4\sec \Theta }{ 16\tan ^2\Theta } 4\sec ^2\Theta d \Theta \]

\[x = 4\tan \Theta \]
\[dx = 4\sec ^2\Theta d \Theta \]

Answer 3

Next I did:

\[\int\limits_{}^{}\frac{ \sec ^3\Theta }{ \tan^2\Theta } d \Theta \]

Answer 4

And now I'm kind of stuck.

Answer 5

It is apparently able to simplify to:

\[\int\limits_{}^{}\tan \Theta d \Theta \]

But I have no idea how to get there.

Answer 6

express top and bottom in terms of sin and cos

Answer 7

Next I did:

\[\int\limits_{}^{}\frac{ \sec ^3\Theta }{ \tan^2\Theta } d \Theta =\int\limits_{}^{}\frac{ 1}{\cos^3\Theta \dfrac{\sin^2\Theta}{\cos^2\Theta}} d \Theta \]

Answer 8

So you have
\[\int\limits_{}^{}\frac{ 1 }{ { \sin^2\Theta \cos \Theta }{ } }\]

Answer 9

whoops forgot the d theta

Answer 10

yeah im also hating trig substitution atm..

Answer 11

\[\csc^2\Theta \sec\Theta = (1+\cot^2\Theta)\sec\Theta = \sec\Theta + \csc\Theta \cot\Theta\]

Answer 12

this is so frustrating

Answer 13

Indeed..

Answer 14

got any other better ideas ?

Answer 15

By parts from the starttt

Answer 16

familiar with Integration By Parts @unimatix ?

Answer 17

Yes

Answer 18

OHH!

Answer 19

I'm guessing this is wrong.
\[u = \tan \Theta \]
\[du = \sec ^2 d\Theta \]

Answer 20

\[\frac{ 1 - u^2 }{ u^2 } du\]

Answer 21

\[\int\limits_{}^{}\frac{ 1 }{ u^2 } - 1 du\]

Answer 22

Does that look right to anyone?

Answer 23

Okay if anyone comes up with anything I'll check in the morning. Thanks for all of your help!

Answer 24

Ok...i come across this....after u converted it into sin n cos...u were left vid integral of 1/cos A sin^2A

Answer 25

Now u can write it as integral of cosec^2 A sec A dA

Answer 26

Open up cosec^2 A as (1+cot^2A)

Answer 27

N u will get integral of(1+cot^2A)secA dA

Answer 28

Open up the brackets n solve it....

Answer 29

U will get the ans aa log|secA +tanA| -cosec A

Answer 30

Now put back the value of A as
arc tan(x/4)