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I'm trying to understand exactly when trigonometric substitution can be used. Here's the whole detailed process and the approach I'm currently taking. I'm mainly curious about whether the trigonometric substitution in step 4 is correct. It appears so to me, but when I use a CAS or graphing system to integrate, its giving me nothing close.

I have this integral:

$\int{\frac{x+4}{x^{2}+2x+5}dx}$

My steps for approaching this are the following:

  1. Complete the square:

$\int{\frac{x+4}{\left(x^{2}+2x+1\right)+5-1}dx} = \int{\frac{x+4}{\left(x+1\right)^{2}+4}dx} = \int{\left(x+4\right)\frac{1}{\left(x+1\right)^{2}+4}dx}$

  1. Assume a triangle of the following sides:

base = $2$, height = $x+1$, hypotenuse = $\sqrt{\left(x+1\right)^{2}+4}$

  1. Use the following for substitutions:

$\cos \left(\theta\right)=\frac{2}{\sqrt{\left( x+1 \right)^{2}+4}}\; so\; \frac{\cos ^{2}\left( \theta \right)}{4}=\frac{1}{\left( x+1 \right)^{2}+4}$

$\tan \left(\theta\right)=\frac{x+1}{2}\; so\; 2\tan\left(\theta\right)-1=x$

...and

$2\sec^{2}\left(\theta\right)d\theta=dx$

  1. Substituting to get the following:

$\int{\left( \left( 2\tan \left( \theta \right)-1 \right)+4 \right)\frac{\cos ^{2}\left( \theta \right)}{4}2\sec ^{2}\left(\theta\right) d\theta }$

  1. Simplifying, I get:

$\int{\left( \tan \left( \theta \right)+\frac{3}{2} \right)d\theta } = \int{\left( \frac{\sin \left( \theta \right)}{\cos\left( \theta \right)}+\frac{3}{2} \right)d\theta }$

  1. Substituting $u=\cos\left(\theta\right)$, and $-du=\sin\left(\theta\right)d\theta$ and integrating, I get...

$-\int{\frac{1}{u}}du+\frac{3}{2}\int{d\theta}$ = $-\ln{\left|u\right|}+\frac{3\theta}{2}+C$

  1. Converting back to $\theta$ I get...

$-\ln{\left|\cos\left(\theta\right)\right|}+\frac{3\theta}{2}+C$

  1. Converting back to x I get the final result:

$\frac{3\arctan\left(\frac{x+1}{2}\right)}{2}-\ln{\left| \frac{2}{\sqrt{x^{2}+2x+5}} \right|}+C$

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    $\begingroup$ There are a number of ways to verify your answer. 1) You can derive your answer and compare with the function to be integrated; this even easier if you have a CAS. 2) If the book gives you another answer, you can plot your answer and the answer from the book; the two functions should be parallel since they may differ by no more than a constant if both are right. Or you can subtract the two answers and simplify; again, you should get a constant, if your answer is right. $\endgroup$ – Bernard Massé Mar 21 '15 at 16:23
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I'm mainly curious about whether the trigonometric substitution in step 4 is correct. It appears so to me, but when I use a CAS or graphing system to integrate, its giving me nothing close.

Your step 4 and your global approach are correct, and your answer written as $$ \frac{3}{2} \arctan\left(\frac{1+x}{2}\right)+\frac{1}{2} \log\left(5+2 x+x^2\right)+C $$ is correct too, since you can easily check that $$ \left(\frac{3}{2} \arctan\left(\frac{1+x}{2}\right)+\frac{1}{2} \log\left(5+2 x+x^2\right)+C\right)'=\frac{x+4}{x^{2}+2x+5}. $$

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  • $\begingroup$ For some stupid reason, the obvious approach to checking seems to evade me. :D Thanks! $\endgroup$ – Topher Mar 21 '15 at 19:45

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