HSC Mathematics Advanced, Extension 1, and Extension 2/3-Unit/HSC/Primitive of sin and cos squared

The integration of these functions can be performed by the substitution of $$\sin^2\alpha$$ or $$\cos^2\alpha$$ by a function in terms of $$\cos2\alpha$$ respectively.

Deriving the substituted functions
By the double-angle formula, $$\left.\cos 2\alpha = 2\cos^2 \alpha - 1\right.$$. Rearranging to make $$\cos^2\alpha$$ the subject, $$\cos^2\alpha = \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha$$.

We know by the Pythagorean identity $$\left.\sin^2\theta + \cos^2\theta = 1\right.$$ that $$\left.\cos^2\alpha = 1 - \sin^2\alpha\right.$$, so $$1 - \sin^2\alpha = \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha$$. Rearranging, $$\sin^2\alpha = \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha$$.

Finding the primitive
By substituting $$\sin^2\alpha\,$$ for $$\tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha$$, $$\int \sin^2 \alpha d\alpha = \int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha d\alpha$$. Similarly for $$\cos^2\alpha \,$$,

$$\int \cos^2 \alpha \, d\alpha = \int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha$$

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

$$ \int \cos \alpha d\alpha = \sin \alpha + c$$

$$ \int \sin \alpha d\alpha = - \cos \alpha + c$$

For cos2 α

$$\int \cos^2 \alpha \, d\alpha = \int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha$$

$$\int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha = \tfrac{1}{2} \alpha + \tfrac{1}{4} \sin2\alpha + c $$

$$\int \cos^2 \alpha \, d\alpha = \tfrac{1}{2} \alpha + \tfrac{1}{4} \sin2\alpha + c $$

For sin2 α

$$\int \sin^2 \alpha \, d\alpha = \int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha \, d\alpha$$

$$\int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha \, d\alpha = \tfrac{1}{2} \alpha - \tfrac{1}{4} \sin2\alpha + c $$

$$\int \sin^2 \alpha \, d\alpha = \tfrac{1}{2} \alpha - \tfrac{1}{4} \sin2\alpha + c $$