Question

How do you integrate `int (e^x)/sqrt(e^(2x) +16)dx` using trigonometric substitution?

Answer 1

`inte^x/sqrt(e^(2x)+16)dx=ln|sqrt(e^(2x)+16)+e^x|+c`

Explanation:

Let `e^x=4tanu` then `e^xdx=4sec^2udu` and

`dx=(4sec^2u)/(4tanu)du` -as `e^x=4tanu`

and `sqrt(e^(2x)+16)=sqrt(16tan^2u+16)=4secu`

Hence `inte^x/sqrt(e^(2x)+16)dx`

= `int(4tanu)/(4secu)(4sec^2u)/(4tanu)du`

= `intsecudu`

= `ln|secu+tanu|+c_1`

= `ln|sqrt(tan^2u+1)+tanu|+c_1`

= `ln|sqrt(e^(2x)+16)/4+e^x/4|+c_1`

= `ln|sqrt(e^(2x)+16)+e^x|+c`, where `c` is another constant.