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can you explain a little better how do you obtain sec^2(t) from 1+cos^2(2x) with the substitution tan(t)=cos(2x), ( i am referring to the denominator in the second line)please?
And can you also explain the reason why the minus sign in the substitution disappear? Should not be [-( sec^2(t)/sec^2(t))]?
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I have added a line and red markings to explain your two questions. Hope that they help.
I = ∫ 0 2 π 3 + 2 cos 2 2 x − 1 4 sin 2 x = − ∫ 0 2 π 1 + cos 2 2 x − 2 sin 2 x k n o w i n g t h a t ( cos 2 x ) ′ = − 2 sin 2 x s o , I = { 2 π 0 − arctan ( cos 2 x ) = 2 π
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∫ 0 2 π 3 + cos ( 4 x ) 8 sin x cos x d x = ∫ 0 2 π 3 + 2 cos 2 ( 2 x ) − 1 4 sin ( 2 x ) d x = ∫ 0 2 π 1 + cos 2 ( 2 x ) 2 sin ( 2 x ) d x Let tan θ = cos ( 2 x ) ⇒ sec 2 θ d θ = − 2 sin ( 2 x ) d x = ∫ 4 π − 4 π 1 + tan 2 θ − sec 2 θ d θ Note that 1 + tan 2 θ = sec 2 θ = ∫ − 4 π 4 π sec 2 θ sec 2 θ d θ = ∫ − 4 π 4 π d θ = [ θ ] − 4 π 4 π = 2 π