The integration of these functions can be performed by the substitution of sin2α{\displaystyle \sin ^{2}\alpha } or cos2α{\displaystyle \cos ^{2}\alpha } by a function in terms of cos2α{\displaystyle \cos 2\alpha } respectively.
By the double-angle formula, cos2α=2cos2α−1{\displaystyle \left.\cos 2\alpha =2\cos ^{2}\alpha -1\right.} . Rearranging to make cos2α{\displaystyle \cos ^{2}\alpha } the subject, cos2α=12+12cos2α{\displaystyle \cos ^{2}\alpha ={\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha } .
We know by the Pythagorean identity sin2θ+cos2θ=1{\displaystyle \left.\sin ^{2}\theta +\cos ^{2}\theta =1\right.} that cos2α=1−sin2α{\displaystyle \left.\cos ^{2}\alpha =1-\sin ^{2}\alpha \right.} , so 1−sin2α=12+12cos2α{\displaystyle 1-\sin ^{2}\alpha ={\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha } . Rearranging, sin2α=12−12cos2α{\displaystyle \sin ^{2}\alpha ={\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha } .
By substituting sin2α{\displaystyle \sin ^{2}\alpha \,} for 12−12cos2α{\displaystyle {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha } , ∫sin2αdα=∫12−12cos2αdα{\displaystyle \int \sin ^{2}\alpha d\alpha =\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha d\alpha } . Similarly for cos2α{\displaystyle \cos ^{2}\alpha \,} ,
∫cos2αdα=∫12+12cos2αdα{\displaystyle \int \cos ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha }
These can be integrated by using the primitives of cosine and sine
∫cosαdα=sinα+c{\displaystyle \int \cos \alpha d\alpha =\sin \alpha +c}
∫sinαdα=−cosα+c{\displaystyle \int \sin \alpha d\alpha =-\cos \alpha +c}
For cos2 α
∫12+12cos2αdα=12α+14sin2α+c{\displaystyle \int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha ={\tfrac {1}{2}}\alpha +{\tfrac {1}{4}}\sin 2\alpha +c}
∫cos2αdα=12α+14sin2α+c{\displaystyle \int \cos ^{2}\alpha \,d\alpha ={\tfrac {1}{2}}\alpha +{\tfrac {1}{4}}\sin 2\alpha +c}
For sin2 α
∫sin2αdα=∫12−12cos2αdα{\displaystyle \int \sin ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha \,d\alpha }
∫12−12cos2αdα=12α−14sin2α+c{\displaystyle \int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha \,d\alpha ={\tfrac {1}{2}}\alpha -{\tfrac {1}{4}}\sin 2\alpha +c}
∫sin2αdα=12α−14sin2α+c{\displaystyle \int \sin ^{2}\alpha \,d\alpha ={\tfrac {1}{2}}\alpha -{\tfrac {1}{4}}\sin 2\alpha +c}
for 
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. Similarly for 
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