# Integrating dx / (4+x^2)^2 using Trigonometric Substitution

## Homework Statement

Evaluate the integral:
integral of dx / (4+x^2)^2

## Homework Equations

x = a tan x theta
a^2 + x^2 = a^2 sec^2 theta

## The Attempt at a Solution

x = 2 tan theta
dx = 2sec^2 theta
tan theta = x/2

integral of dx / (4+x^2)^2
= 1/8 integral (sec^2 theta / sec^4 theta) d theta
= 1/8 integral (cos^2 theta) dtheta
= 1/16 integral 1 + cos 2 theta d theta
= 1/16 [ theta + sin 2 theta / 2 ] + c
= 1/16 [ theta + sin theta cos theta ] + c
= 1/16 [ tan ^-1 x/2 + ??? ] + c

"???" is where I get lost.
Did I do the problem wrong? I was really confused when the book gave me a problem without a square root.

## Answers and Replies

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Student100
Gold Member

## Homework Statement

Evaluate the integral:
integral of dx / (4+x^2)^2

## Homework Equations

x = a tan x theta
a^2 + x^2 = a^2 sec^2 theta

## The Attempt at a Solution

x = 2 tan theta
dx = 2sec^2 theta
tan theta = x/2

integral of dx / (4+x^2)^2
= 1/8 integral (sec^2 theta / sec^4 theta) d theta
= 1/8 integral (cos^2 theta) dtheta
= 1/16 integral 1 + cos 2 theta d theta
= 1/16 [ theta + sin 2 theta / 2 ] + c
= 1/16 [ theta + sin theta cos theta ] + c
= 1/16 [ tan ^-1 x/2 + ??? ] + c

"???" is where I get lost.
Did I do the problem wrong? I was really confused when the book gave me a problem without a square root.
Your sub looks fine, and your work up to $$=\frac{\theta}{16}+\frac{sin(2\theta)}{32}+C$$

Go ahead and make your last back sub.

a^2 + x^2 = a^2 sec^2 theta
x^2 + 4 = 4 sec^2 theta
sqrt (x^2 + 4) / 2 = sec theta
cos sqrt (x^2 + 4) / 2 = theta

Is this what you mean by back sub?

Student100
Gold Member
a^2 + x^2 = a^2 sec^2 theta
x^2 + 4 = 4 sec^2 theta
sqrt (x^2 + 4) / 2 = sec theta
cos sqrt (x^2 + 4) / 2 = theta

Is this what you mean by back sub?
You have ##\theta = tan^{-1}(\frac{x}{2})## above, just plug in for theta and simplify.

So,

= [ θ/16 + sin (2θ)/32 ] + C
= [ tan^-1 (x/2) / 16 + sin (2 tan^-1 (x/2) / 32 ] + C

Is it okay to plug in tan^-1 (x/2) for θ in sin (2θ) or am I misinterpreting you?
It seems weird.

Student100
Gold Member
So,

= [ θ/16 + sin (2θ)/32 ] + C
= [ tan^-1 (x/2) / 16 + sin (2 tan^-1 (x/2) / 32 ] + C

Is it okay to plug in tan^-1 (x/2) for θ in sin (2θ) or am I misinterpreting you?
It seems weird.
It's perfectly fine, that's what you needed to do. You had already found above that ##tan(\theta)=\frac{x}{2}##, you could simplify the sin by redoing the double angle formula ##sin(2tan^{-1}(\frac{x}{2}))=2(sin(tan^{-1}(\frac{x}{2}))(cos(tan^{-1}(\frac{x}{2}))## and simplify from your right triangle picture.

Why does it seem weird?

= [ tan^-1 (x/2) / 16 + (sin(4/x) cos(4/x))/16 ]

Does it simplify to that?

It seems weird because I don't think I've done it before and the answer in the back of the book is in a different format:
(tan^-1 (x/2) )/ 16 + x/(8(4 + x^2)

Sorry for the messy and inconsistent typing, and thanks for the help.

Student100
Gold Member
= [ tan^-1 (x/2) / 16 + (sin(4/x) cos(4/x))/16 ]

Does it simplify to that?

It seems weird because I don't think I've done it before and the answer in the back of the book is in a different format:
(tan^-1 (x/2) )/ 16 + x/(8(4 + x^2)

Sorry for the messy and inconsistent typing, and thanks for the help.
No problem, so you havehttps://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/73/73867-db7eccc9bcdd5e7324f0dfba67853800.jpg [Broken]
$$= \frac{1}{16}(tan^{-1}(\frac{x}{2})+(sin(tan^{-1}(\frac{x}{2}))(cos(tan^{-1}(\frac{x}{2})))$$

Since $$\theta=tan^{-1}(\frac{x}{2})$$
$$cos(tan^{-1}(\frac{x}{2}))=\frac{2}{\sqrt{x^2+4)}}$$

Do the same for sin, simplify, and you should get the book answer.

Last edited by a moderator:
I did it! Looks exactly like the book's.

Thanks again.

Student100