Integral question (1 Viewer)

Axio · Mar 19, 2015

Hi guys, can anyone help me with this integral?

$I=\int \sqrt{16+x^2}dx$

I know that there would be easier ways to solve it but can anyone see where I've gone wrong?

I use a trigonometric substitution and then integration by parts. Assuming my other steps are right, I get to:

$\frac{I}{8}=tan\theta sec\theta +ln|tan\theta +sec\theta| +c$

And then by substituting in:

$$Substituting in$ \ x=4tan\theta$

I get:

$I=\frac{x}{2}\sqrt{x^2 +16}+8ln|\sqrt{\frac{x^2}{16}+1}+\frac{x}{4}|+c$

... Which doesn't seem to be the right answer :/.

EDIT:

I think my mistake is coming from substituting it back in...

integral95 · Mar 19, 2015

Axio said:
Hi guys, can anyone help me with this integral?

$I=\int \sqrt{16+x^2}dx$

I know that there would be easier ways to solve it but can anyone see where I've gone wrong?

I use a trigonometric substitution and then partial fractions. Assuming my other steps are right, I get to:

$\frac{I}{8}=tan\theta sec\theta +ln|tan\theta +sec\theta| +c$

And then by substituting in:

$$Substituting in$ \ x=4tan\theta$

I get:

$I=\frac{x}{2}\sqrt{x^2 +16}+8ln|\sqrt{\frac{x^2}{16}+1}+\frac{x}{4}|+c$

... Which doesn't seem to be the right answer :/.

EDIT:

I think my mistake is coming from substituting it back in...

What trig substitution made you use partial fraction afterwards?

$$if you let$ \ \ x = 4\tan\theta \\ \\ \Rightarrow 16 \int sec^3 \theta \ \ d\theta$

From there you use IBP (integration by parts) then you sub it back in.

(i dunno if that's the best method though lol)

Axio · Mar 20, 2015

Lol sorry I meant integration by parts... What does sec(tan^-1(x/4)) equal?

astroman · Mar 20, 2015

http://www.wolframalpha.com/input/?i=sec(tan^-1(x/4))

InteGrand · Mar 20, 2015

Axio said:
Lol sorry I meant integration by parts... What does sec(tan^-1(x/4)) equal?

(Use sec² t = 1 + tan² t)
$\sec\left ( \tan ^{-1}\frac{x}{4} \right )=\sqrt{1+\tan ^2 \left ( \tan ^{-1}\frac{x}{4} \right )}=\sqrt{1+\left ( \frac{x}{4} \right )^2}=\frac{1}{4}\sqrt{16+x^2}$ (taking positive square root since sec is positive when the argument is between $-\frac{\pi}{2}$ and $+\frac{\pi}{2}$ , which it is here, since the argument is an inverse tan)

Integral question (1 Viewer)

Axio

=o

integral95

Well-Known Member

Axio

=o

astroman

Well-Known Member

InteGrand

Well-Known Member

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