Francisca Rodden

2021-12-28

Can trigonometric substitution be used to solve this integral?

The integral in question is

I first completed the square in the denominator giving

It is at this point I used trigonometric substitution with

which is incorrect. The correct answer is

What went wrong with my trig substitution?

Paineow

Beginner2021-12-29Added 30 answers

Since $u=2\mathrm{tan}\theta$ , you have

${u}^{2}=4{\mathrm{tan}}^{2}\theta =4({\mathrm{sec}}^{2}\theta -1)$

So,

$\mathrm{sec}\theta =\sqrt{\frac{{u}^{2}+4}{4}}$

and therefore

$\frac{19}{\mathrm{ln}\left(\mathrm{sec}\theta \right)}=\frac{1}{18}\mathrm{ln}\left(\frac{{u}^{2}+4}{4}\right)$

So,

and therefore

alexandrebaud43

Beginner2021-12-30Added 36 answers

There is a simpler way to determine this integral: rewrite first the numerator:

$x+1=\frac{1}{18}(18x+6)+\frac{23}{}$

split the integral in two:

$\int \frac{x+1}{9{x}^{2}+6x+5}dx=\frac{1}{18}\int \frac{18x+6}{9{x}^{2}+6x+5}dx+\frac{23}{\int}\frac{dx}{{(3x+1)}^{2}+4}$

$=\frac{1}{18}\mathrm{ln}(9{x}^{2}+6x+5)+\frac{23}{\int}\frac{dx}{{(3x+1)}^{2}+4}$

$=\frac{1}{18}\mathrm{ln}(9{x}^{2}+6x+5)+\frac{29}{\int}\frac{d(3x+1)}{{(3x+1)}^{2}+4}$

and use the standard formula

$\int \frac{dx}{{x}^{2}+{a}^{2}}=\frac{1}{a}\mathrm{arctan}\left(\frac{x}{a}\right)$

split the integral in two:

and use the standard formula

nick1337

Expert2022-01-08Added 777 answers

Which for the above:

22+64

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