Question

Find integral sign 10/(4+x^2) dx using the substitution x=2 tan theta. (Give your answer in terms of x.)

Answer 1

\[10\int\frac{1}{4+x^2}dx\]

Answer 2

yes

Answer 3

not sure why you need this substitution, since the derivative of
\[\tan^{-1}(x)=\frac{1}{1+x^2}\] so you are close already, but we can do it

Answer 4

so put
\[x=2\tan(\theta)\]
\[dx=2\sec^2(\theta)d\theta\] and so you get
\[\int\frac{\sec^2(\theta)}{4+4\tan^2(\theta)}d\theta\]

Answer 5

denominator is
\[4(1+\tan^2(\theta))=4\sec^2(\theta)\]

Answer 6

so now your entire integral is
\[\frac{1}{4}\int\frac{sec^2(\theta)}{\sec^2(\theta)}d\theta\]

Answer 7

in other words just
\[\frac{1}{4}\int1d\theta\]

Answer 8

anti-derivative is
\[\frac{1}{4}\theta\]

Answer 9

now we have to go back.
\[x=2\tan(\theta)\iff \frac{x}{2}=\tan(\theta)\iff \theta=\tan^{-1}(\frac{x}{2})\]

Answer 10

hmm i seem to be off by a factor of 2

Answer 11

oh i am an idiot.
\[x=2\tan(\theta)\]
\[dx=2\sec^2(\theta)d\theta\] that is why. there should be another 2 out front!

Answer 12

so stick a 2 out front and cancel to get
\[\int\frac{1}{4+x^2}dx=\frac{1}{2}\tan^{-1}(\frac{x}{2})\]

Answer 13

i guess you have to multiply by 10 so "final answer" is
\[5\tan^{-1}(\frac{x}{2})+C\]

Answer 14

sorry it took a while but these stupid notifications are blocking my view

Answer 15

thanx

Answer 16

yw hope steps are clear and sorry i messed up with the 2