How do you evaluate the definite integral #int(2/sqrt(1-x^2)+e^(2x))dx# from #[0,1]#?

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Answer 1 · 2017-09-20T05:29:25

The answer is #=pi+1/2(e^2-1)=6.33612...#

There are 2 integrals to calculate

#int_0^1(2/sqrt(1-x^2)+e^(2x))dx=int_0^1 2/sqrt(1-x^2)dx+int_0^1e^(2x)dx#

Perform the first integral by substitution

Let

#x=sintheta#, #=>#, #dx=costhetad theta#

#sqrt(1-x^2)=sqrt(1-sin^2theta)=costheta#

Therefore,

The first integral is

#int_0^1 2/sqrt(1-x^2)dx=int_0^(pi/2)(2costhetad theta)/(costheta)#

#=[2theta]_0^(pi/2)=pi#

The second integral is

#int_0^1e^(2x)dx=[e^(2x)/2]_0^1=(e^2/2)-(1/2)#

So,

Putting it al ltogether,

#int_0^1(2/sqrt(1-x^2)+e^(2x))dx=pi+1/2(e^2-1)#