Question

integral of 1/(x^2 * sqrt(x^2 + 4)) where x=2tan(theta). I've got it to 1/4 * integral of 1/tan(theta)^2 d(theta) .... not sure where to from there...

Answer 1

dx/ (x^2 * sqrt(x^2 + 4)) and you substituted x= tan@ right ?

Answer 2

2tan@ *

Answer 3

??

Answer 4

Your substitution is correct, just don't forget to compute the derivative of it too, and back substitute for all given terms of x.

Answer 5

How do i integrate \[\int\limits_{}^{} {\frac{ 1 }{ \tan(\theta)^2}}\] ?

Answer 6

is that what you end up with after back substitution ?

Answer 7

i get 1/4 * (what i wrote up there) before substituting back...which i can only do after integrating that

Answer 8

\[ \large x= 2 \tan \alpha \longrightarrow \frac{dx}{d\alpha}= 2 \sec^2 \alpha \]

\[\int \frac{2 \sec^2 \alpha}{4\tan^2 \alpha2 \cdot \sec\alpha}=\frac{1}{4} \int \frac{\sec \alpha}{\tan ^2 \alpha } = \frac{1}{4}\int \frac{\cos \alpha}{\sin ^2 \alpha}d\alpha \]

Answer 9

Or did I make a mistake there?

Answer 10

seems right so far.

Answer 11

follow that @remnant ?

Answer 12

yeah. sorry...just saw my mistake...trying it again quickly

Answer 13

great, the integral above is easy to compute. Give it a try.

Answer 14

for the last integral there do I use substitution where u = sin(a) and du/da = cos(a)?

Answer 15

yes I recommend you doing that, since the numerator in the derivative of the denominator.

Answer 16

then i get \[-\frac{ 1 }{ 4 } u^{-1}\] after integrating. Then subing back in, i get:\[-\frac{ 1 }{ \sin (\arctan (\frac{ x }{ 2 })) }\] which is not the right answer

Answer 17

it is not because you didn't back substitute.

Answer 18

you obtain the following result.

\[ \Large -\frac{1}{4u}\]

and u=sin alpha.

Answer 19

\[\Large x=2\tan \alpha \longrightarrow \tan \alpha = \frac{x}{2}= \frac{opposite}{ank} \]