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Freshman math definite integral?
3)x = sinu; dx = cosu du

Original integral = ∫ [pi/4, pi/2] cos 2u/sin 2udu

=∫[π/4, π/2] CSC 2u-1du

= (-cotu- 1)|[pi/4,pi/2]

= 1-pi/4

10) original integral = ∫ [-pi/2, pi/2] (1+cos2x) 2dx.

= ∫[-pi/2,pi/2]( 1+2 cos2x+( 1/2)( 1+cos4x))dx

= 3pi/2

12) original integral = ∫ [- 1, 1] (arctanx) 2 d arctanx, (att: odd function symmetric area integral is zero).

=( 1/3)(arctanx)^3|[- 1, 1)

= pi^3/96