Question

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

Answer 1

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

Explanation:

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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• Since `d/(dx) sec theta = secthetatantheta`

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

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

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