The integration of these functions can be performed by the substitution of ${\displaystyle \sin ^{2}\alpha }$ or ${\displaystyle \cos ^{2}\alpha }$ by a function in terms of ${\displaystyle \cos 2\alpha }$ respectively.

Deriving the substituted functions

By the double-angle formula, ${\displaystyle \left.\cos 2\alpha =2\cos ^{2}\alpha -1\right.}$. Rearranging to make ${\displaystyle \cos ^{2}\alpha }$ the subject, ${\displaystyle \cos ^{2}\alpha ={\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha }$.

We know by the Pythagorean identity ${\displaystyle \left.\sin ^{2}\theta +\cos ^{2}\theta =1\right.}$ that ${\displaystyle \left.\cos ^{2}\alpha =1-\sin ^{2}\alpha \right.}$, so ${\displaystyle 1-\sin ^{2}\alpha ={\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha }$. Rearranging, ${\displaystyle \sin ^{2}\alpha ={\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha }$.

Finding the primitive

By substituting ${\displaystyle \sin ^{2}\alpha \,}$ for ${\displaystyle {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha }$, ${\displaystyle \int \sin ^{2}\alpha d\alpha =\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha d\alpha }$. Similarly for ${\displaystyle \cos ^{2}\alpha \,}$,

${\displaystyle \int \cos ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha }$

These can be integrated by using the primitives of cosine and sine

${\displaystyle \int \cos \alpha d\alpha =\sin \alpha +c}$

${\displaystyle \int \sin \alpha d\alpha =-\cos \alpha +c}$

For cos² α

${\displaystyle \int \cos ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha }$

${\displaystyle \int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha ={\tfrac {1}{2}}\alpha +{\tfrac {1}{4}}\sin 2\alpha +c}$

${\displaystyle \int \cos ^{2}\alpha \,d\alpha ={\tfrac {1}{2}}\alpha +{\tfrac {1}{4}}\sin 2\alpha +c}$

For sin² α

${\displaystyle \int \sin ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha \,d\alpha }$

${\displaystyle \int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha \,d\alpha ={\tfrac {1}{2}}\alpha -{\tfrac {1}{4}}\sin 2\alpha +c}$

${\displaystyle \int \sin ^{2}\alpha \,d\alpha ={\tfrac {1}{2}}\alpha -{\tfrac {1}{4}}\sin 2\alpha +c}$