How do you integrate $\int \frac{1}{\sqrt{x^{2} + 4}}$ $int 1/sqrt(x^2+4)$ by trigonometric substitution?

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Answer 1 · 2016-07-25T21:08:56

$\int \frac{d x}{\sqrt{x^{2} + 4}} = l n (\frac{x}{2} + \sqrt{1 + \frac{x^{2}}{4}}) + C$ $int (d x)/(sqrt (x^2+4))=l n(x/2+sqrt(1+x^2/4))+C$

$\int \frac{d x}{\sqrt{x^{2} + 4}} = ?$ $int (d x)/(sqrt (x^2+4))=?$

$let be u = \frac{x}{2}; d x = 2 \cdot d u; x^{2} = 4 u^{2}$ $"let be "u=x/2" ; " d x=2*d u" ; "x^2=4u^2$

$\int \frac{2 \cdot d u}{\sqrt{4 u^{2} + 4}} = \int \frac{2 \cdot d u}{\sqrt{4 (u^{2} + 1)}}$ $int (2*d u)/(sqrt(4u^2+4))=int(2*d u)/(sqrt(4(u^2+1))$

$int(cancel(2)* d u)/(cancel(2)*sqrt(u^2+1))=int(d u)/(sqrt(u^2+1))$

$"now, substitute "u=tan v" ; "v=arc tan u$

$d u=sec ^2 v* d v$

$int (sec ^2 v*d v)/(sqrt(tan^2 v+1))" ;so "tan^2 v +1=sec ^2 v$

$int(sec^2 v*d v)/(sqrt(sec^2 v))=int (cancel(sec)^2 v*d v)/(cancel(sec) v)=int sec v*d v$

$"expand fraction by " tan v+sec v$

$int sec v*d v*(tan v+sec v)/(tan v+sec v)$

$int (sec v*tan v + sec ^2 v)/(tan v + sec v) *d v$

$k=tan v+ sec v$

$d k=(sec v* tan v+sec ^2 v) *d v$

$int ( d k)/k=l n k+C$

$"undo substitution "k=tan v +sec v$

$int (d x)/(sqrt (x^2+4))= l n(tan v+sec v)+C$

$sec v=sqrt(1+tan ^2 v)=sqrt (1+u^2)$

$int (d x)/(sqrt (x^2+4))=l n(u+sqrt(1+u^2))+C$

$"undo substitution " u=x/2$

$int (d x)/(sqrt (x^2+4))=l n(x/2+sqrt(1+x^2/4))+C$

How do you integrate int 1/sqrt(x^2+4)∫1√x2+4int 1/sqrt(x^2+4) by trigonometric substitution?