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

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Answer 1 · 2018-03-14T06:05:13

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

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.