What is the integral of (1+sinx)^(1/2)dx?

Question

The answer that I need to have it as is -2cos(x)/(1+sin(x))^(1/2).  How would I go about getting this as the answer from the integral?

? · Accepted Answer

Not sure of the easiest way but this is one way to do it

                    Multiply the integrand by 1 in the form of (1 - sin x)^(1/2)/ (1 - sin x)^(1/2)

                    the integrand is now   (1 + sinx)^(1/2)(1 - sin x)^(1/2)/ (1 - sin x)^(1/2) 
                                                       = (1 - sin^2x)^(1/2)/ (1 - sin x)^(1/2)
                                                       = (cos^2 x)^(1/2)/(1 - sin x)^(1/2)
                                                      =  cos x/(1 - sin x)^(1/2)

Integrate by letting u = (1 - sin x)  and du = - cos x dx

                                          integral -du/u^(1/2) = -2u^(1/2) + C

Resubstitute                            -2(1 - sin x)^(1/2) + c

rationalize the numerator           -2(1 - sin x)^(1/2)(1 + sin x)^(1/2)/(1 + sin x)^(1/2)
                                               -2 (1 - sin^2 x)^(1/2)/(1 + sin x)^(1/2)
                                               -2 cos x/(1 + sin x)^(1/2) + c

Anonymous · Answer

wolframalpha.com

search it at google.

that site will give every single equations.

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What is the integral of (1+sinx)^(1/2)dx?

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