How do you integrate #int xsqrt(x^2+4)# by trigonometric substitution?

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Answer 1 · 2018-04-30T01:03:56

#(8(sec theta)^3)/3 + C#

Let x=#2tantheta#
#(dx)/(d theta)=2(sec theta)^2#
#dx=2(sec theta)^2 (d theta)#

#int xsqrt(x^2+4) dx#

=#int 2tan theta sqrt((2tan theta)^2+4) times 2(sec theta)^2 (d theta)#

=#int2tan theta sqrt(4(tan theta)^2+4) times 2(sec theta)^2 d theta#

=#int2tan thetatimes2sec theta times 2(sec theta)^2 d theta#

=#8inttantheta (sec theta)^3 d theta#

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=#8int(sec theta)^2 times (tanthetasectheta)#

=#8 times(sec theta)^3/3 + C#

=#(8(sec theta)^3)/3 + C#