`int_0^(pi/2) (2 - sin(theta))^2 d theta` Evaluate the integral

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lemjay eNotes educator| Certified Educator

`int_0^(pi/2) (2-sin (theta))^2 d theta`

First, expand the integrand.

`=int_0^(pi/2) (2-sin (theta))(2-sin (theta)) d theta`

`= int _0^(pi/2) (4-4sin (theta)+sin^2 (theta)) d theta`

For the last term of the integrand, apply the trigonometric identity `cos(2theta) = 1 - 2sin^2 (theta)` .

`= int_0^(pi/2) (4-4sin(theta) + 1/2 - cos(2theta)/2) d theta`

`= int_0^(pi/2) (9/2-4sin theta -(cos(2theta))/2) d theta`

`= int_0^(pi/2) 9/2 d theta - int_0^(pi/2) 4sin theta d theta - int_0^(pi/2) (cos(2 theta))/2 d theta`

Apply the integral formulas `int adx = ax`  , `int sin xdx = -cosx`  and `int cos x dx = sinx` .

`= (9/2theta + 4cos(theta) - (sin (2theta))/4 )|_0^(pi/2)`

`= (9/2*pi/2 + 4cos(pi/2) - sin (2*pi/2)/4) - (9/2*0+4cos(0) - (sin (2*0))/4)`

`=((9pi)/4+4*0+0/4)-(0+4*1-0/4)`

`=(9pi)/4 - 4`

 

Therefore,  `int _0^(pi/2) (2-sin(theta))^2 d theta = (9pi)/4 - 4` .