[SPLIT] int [x / (x^2 + 4)] dx using trig idents

[warwick]

Yeah, that was my bad. I figured it out last night looking at the relevant csc/cot identity. I'm a bit stuck on this one.

Solve using trig identities.

\(\displaystyle \L\\\int{(\frac{x}{x^{2}+4}})\)

I'm not going to lie. I'm exhausted. I just got back from Austin (3 hours away) picking up my 1998 Trans Am.

 

[galactus]

Just like before, make the subs:

\(\displaystyle \L\\x=2tan({\theta}), \;\ dx=2sec({\theta})tan({\theta})d{\theta}\)

 

[skeeter]

Yeah, that was my bad. I figured it out last night looking at the relevant csc/cot identity. I'm a bit stuck on this one.

Solve using trig identities.

\(\displaystyle \L\\\int{(\frac{x}{x^{2}+4}})\)

I'm not going to lie. I'm exhausted. I just got back from Austin (3 hours away) picking up my 1998 Trans Am.

why use trig substitution? work smart ... not hard.

\(\displaystyle \L \int \frac{x}{x^{2}+4} dx = \frac{1}{2} \int \frac {2x}{x^2+4} dx = \ln{\sqrt{x^2+4}} + C\)

 

[warwick]

Yeah, that was my bad. I figured it out last night looking at the relevant csc/cot identity. I'm a bit stuck on this one.

Solve using trig identities.

\(\displaystyle \L\\\int{(\frac{x}{x^{2}+4}})\)

I'm not going to lie. I'm exhausted. I just got back from Austin (3 hours away) picking up my 1998 Trans Am.

why use trig substitution? work smart ... not hard.

\(\displaystyle \L \int \frac{x}{x^{2}+4} dx = \frac{1}{2} \int \frac {2x}{x^2+4} dx = \ln{\sqrt{x^2+4}} + C\)

That was the first part. The second was to use trig identities to solve.

 

[soroban]

Hello, warwick!

It sounds like you've never done Trig Substitution before.
Okay, here's a walk-through (with baby steps).

I hope you can do similar problems after you've seen the procedure . . .

Solve using trig identities: .\(\displaystyle \L \int\frac{x}{x^2\,+\,4}\,dx\)

Let: \(\displaystyle \,x \:=\:2\tan\theta\;\;\Rightarrow\;\;dx \:=\:2\sec^2\theta\,d\theta\)

Then: \(\displaystyle \:x^2\,+\,4 \:=\2\tan\theta)^2\,+\,4 \;=\;4\tan^2\theta\,+\,4 \;=\;4(\tan^2\theta\,+\,1) \;=\;4\sec^2\theta\)

Substiitute: \(\displaystyle \L\:\int\frac{2\tan\theta}{4\sec^2\theta}\left(2\sec^2\theta\,d\theta\right) \;=\;\int\tan\theta\,d\theta \;=\;\ln|\sec\theta|\,+\,c\;\) [1]


We must back-substitute.

We had: \(\displaystyle \:x \:=\:2\tan\theta\;\;\Rightarrow\;\;\tan\theta\:=\:\frac{x}{2}\:=\:\frac{opp}{adj}\)

\(\displaystyle \theta\) is in a right triangle with: \(\displaystyle \,opp \:=\:x,\;adj\:=\:2\)
. . Using Pythagorus, we find that: \(\displaystyle \:hyp \:=\:\sqrt{x^2\,+\,4}\)
Hence: \(\displaystyle \:\sec\theta \:=\:\frac{hyp}{adj} \:=\:\frac{\sqrt{x^2\,+\,4}}{2}\)

Substitute into [1]: \(\displaystyle \L\:\ln\left(\frac{sqrt{x^2+4}}{2}\right)\,+\,c \;= \;\ln\left[\frac{\left(x^2\,+\,4\right)^{\frac{1}{2}}}{2}\right] + c\)

. . \(\displaystyle \L= \;\ln(x^2\,+\,4)^{\frac{1}{2}} \,-\,\underbrace{\ln2\,+\,c}_{\text{a constant}}\)

. . \(\displaystyle \L=\;\frac{1}{2}\ln(x^2\,+\,4)\,+\,C\)

 

[warwick]

It sounds like you've never done Trig Substitution before....

Well, I've done them successfully, but they were always under a radical in the original integral. I don't know why it threw me off.

 

[galactus]

I was under the impression you had to use trig sub. As Skeeter said, there is an easier way with this one.

\(\displaystyle \L\\\int\frac{x}{x^{2}+4}dx\)

Let \(\displaystyle \L\\u=x^{2}+4, \;\ du=2xdx, \;\ \frac{du}{2}=xdx\)

\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{u}du\)

\(\displaystyle \L\\\frac{1}{2}ln(u)+C\)

\(\displaystyle \L\\\frac{1}{2}ln(x^{2}+4)+C\)

 

[warwick]

I was under the impression you had to use trig sub. As Skeeter said, there is an easier way with this one....

Your assumption was correct. I had to find the same integral using u-sub and trig identities. I wasn't sure if I could apply trig identities since the expression wasn't under a radical. In my book, they have a table specifying which trig functions to use and so forth. Brain fart.