How do you Integrate ?

Question

#int_0^(2x)sqrt(1+sin(x/2))dx#

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Answer 1 · 2018-05-04T19:04:59

#sin(x/2)+cos(x/2)-1#

#int_0^(2x)sqrt(1+sin(x/2))*dx#

#color(green)(sin(x/2)=2sin(x/4)cos(x/4)#

#color(green)(1=sin^2(x/4)+cos^2(x/4)#

Substitute

#int_0^(2x)sqrt(1+sin(x/2))*dx#

#=int_0^(2x)sqrt(sin^2(x/4)+2sin(x/4)cos(x/4)+cos^2(x/4))*dx#

By Completing The Square

#=int_0^(2x)sqrt((sin(x/4)+cos(x/4))^2)*dx#

#=int_0^(2x)(sin(x/4)+cos(x/4))*dx#

#=[sin(x/4)+cos(x/4)]_0^(2x)#

#=sin(x/2)+cos(x/2)-1#

#color(blue)("if this question was like this " sqrt(1+cos(x/2)) " it can be simplified to look like this" #

#color(blue)(sqrt2*cos(x/4) "using the Double angles Formulae"#