DEFINITE INTEGRAL USING CHANGE OF VARAIBLES

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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