How do you integrate $\int \frac{1}{\sqrt{x^{2} + 1}}$ $int 1/sqrt(x^2+1)$ by trigonometric substitution?

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Answer 1 · 2016-08-14T19:36:04

$ln ∣ ∣ x + \sqrt{x^{2} + 1} ∣ ∣ + C$ $ln|x+sqrt(x^2+1)|+C$ .

We take subst. $x = tan t \Rightarrow d x = {sec}^{2} t d t$ $x=tant rArr dx= sec^2tdt$

Also, $\frac{1}{\sqrt{x^{2} + 1}} = \frac{1}{\sqrt{{tan}^{t} + 1}} = \frac{1}{sec t} = cos t$ $1/sqrt(x^2+1)=1/sqrt(tan^t+1)=1/sect =cost$

Hence, $I = \int \frac{1}{\sqrt{x^{2} + 1}} d t$ $I=int1/sqrt(x^2+1)dt$

$= \int cos t {sec}^{2} t d t$ $=intcostsec^2tdt$

$= \int sec t d t$ $=intsectdt$

$= ln | sec t + tan t |$ $=ln|sect+tant|$

$= ln ∣ ∣ x + \sqrt{x^{2} + 1} ∣ ∣ + C$ $=ln|x+sqrt(x^2+1)|+C$ .

How do you integrate ∫1√x2+1int 1/sqrt(x^2+1) by trigonometric substitution?