Problem 1 :
Solution :
Let t = sin2u dt = 2 sinu cosu dt = sin 2u du |
Let t = cos2u dt = 2 cosu (-sinu) dt = -sin 2u du |
Using partial derivative, we get
u = u, du = du, dv = sin 2u ==> v = -cos 2u/2
∫udv = uv-∫vdu
= u(-cos 2u/2)-∫(-cos 2u/2) du
= u(-cos 2u/2)+(sin 2u/4)
By applying the limits, we get
= π/2((-cosπ)/2)) + (sinπ)/4
= π/2(1/2) + 0
= π/4
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