Question

Integrate dx/(4+x²)½ using trigonometric substitution

Answer 1

Integrate d x/(4+x²)½ using trigonometric substitution

Suppose $tan\theta={x \above{2pt} 2}$

by applying cross multiplication

x=2$tan\theta$

$dx=2sec^2\theta$

$\int {dx\above{2pt}\sqrt{x^2+4} }$

$=\int{2sec^2 \theta\above{2pt}\sqrt{4tan^2\theta+4} } d\theta$

$=\int{2sec^2\theta \above{2pt}\sqrt{4(tan^2\theta+1)} } d\theta$

using trigonometric identity $tan^2\theta+1=sec^2\theta$

$=$ $\int{2sec^2\theta\above{2pt}\sqrt{4(sec^2\theta)} }d\theta$

$=\int{2sec^2\theta\above{2pt} 2sec\theta} d\theta$

$=$ $\int sec\theta d\theta$

multiplying and divided both sides by $sec\theta+tan\theta$

$=\int sec\theta({sec\theta+tan\theta \above{2pt} sec\theta+tan\theta}) d\theta$

$=\int{ sec^2\theta+tan\theta sec\theta\above{2pt} sec\theta+tan\theta}d\theta$

Substitute u=$sec\theta+tan\theta$ (A)

$du=(sec\theta tan\theta+sec^2\theta)d\theta$

$=\int {du \above{2pt} u}$

$=ln|u|+C$

By putting the value of u

$=ln|sec\theta+tan\theta|+C$ (1)

Replace $\theta$ terms with x

$tan\theta$ $={x \above{2pt} 2}$

taking square on both sides

$tan^2\theta=({x \above{2pt} 2})^2$

$sec^2\theta-1={x^2 \above{2pt} 4}$

$sec^2\theta={x^4\above{2pt} 4}+1$

$sec^2\theta={x^2+4 \above{2pt} 4}$

taking square root on both sides

$sec\theta={ \sqrt{x^2+4}\above{2pt} 2}$

By putting the value in equation no 1

$=ln( { \sqrt{x^2+4} \above{2pt} 2}+{x\above{2pt} 2} )+C$ Answer