How do you integrate the following function?

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Answer 1 · 2018-06-13T17:27:05

$- (pi sin (1 text{ rad}))/2 = - 1.3217795320407280894430 cdots$

$2t = u => du = 2 * dt => dt = (du)/2$

$I= int_{0}^{pi} 2 sin (u - 1) cos u (du)/2$

$sin(u - 1) = sin u cos 1 - sin 1 cos u$

$I= int_{0}^{pi} (cos 1 sin u cos u - sin 1 cos^2 u) text{ d}u$

$cos^2 x - sin^2 x = cos 2x$
$cos^2 x + sin^2 x = 1$

$I= cos 1 int_{0}^{pi} 1/2 sin 2u text{ d}u - sin 1 int_0^{pi} 1/2 (1 + cos 2u) text{ d}u$

$2u = w => dw = 2 * du => du = (dw)/2$

$I= 1/2 cos 1 int_{0}^{2pi} sin w (dw)/2 - 1/2 sin 1 [pi - 0 + int_0^{2pi} cos w (dw)/2]$

$I= 1/4 cos 1 int_{0}^{2pi} sin w text{ d}w - pi/2 sin1 - 1/4 sin 1 int_0^{2pi} cos w text{ d}w$

$I= 1/4 cos 1 (0 - 0) - pi/2 sin1 - 1/4 sin 1 (1-1)$

$I= - (pi sin1)/2$