Trigonometric Integrals

Definition

Trigonometric integrals are integrals involving powers of sine and cosine. Trigonometric substitution is a technique for integrals containing square roots of quadratic expressions, using inverse substitutions to convert them into trigonometric integrals.

Intuition

Powers of sine and cosine interact through the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ and the half-angle formulas. These identities let us reduce high powers to lower ones or swap between the two functions, creating something amenable to substitution. For square-root integrands, a trigonometric substitution works because a trig identity collapses the square root: $\sqrt{a^2-x^2}$ becomes $a\cos \theta$ when $x=a\sin \theta$, eliminating the radical entirely.

Formal Description

Trigonometric integrals of the form $\int {\sin }^m\theta {\cos }^n\theta d\theta$:

• If both $m$ and $n$ are even, apply the half-angle (reduction) formulas:

$${\sin }^2\theta =\frac{1}{2}(1-\cos 2\theta ),{\cos }^2\theta =\frac{1}{2}(1+\cos 2\theta ).$$

• If either $m$ or $n$ is odd, apply the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ to reduce to a single trigonometric function, then use substitution.

Trigonometric substitution handles integrands containing:

$$\sqrt{a^2-x^2},\sqrt{a^2+x^2},\sqrt{x^2-a^2}.$$

Use the respective inverse substitutions:

$$x=a\sin \theta ,x=a\tan \theta ,x=a\sec \theta ,$$

then find $dx$ by differentiation and change the limits of integration accordingly.

Example 1 (even powers). ${\int }_0^{2\pi }{\cos }^2\theta d\theta$:

$${\int }_0^{2\pi }{\cos }^2\theta d\theta =\frac{1}{2}{\int }_0^{2\pi }(1+\cos 2\theta )d\theta =\frac{1}{2}{\left[\theta +\frac{1}{2}\sin 2\theta \right]}_0^{2\pi }=\pi .$$

Example 2 (odd power). ${\int }_0^{\pi /2}{\cos }^3\theta d\theta$. Write ${\cos }^3\theta =(1-{\sin }^2\theta )\cos \theta$ and let $u=\sin \theta$, $du=\cos \theta d\theta$:

$${\int }_0^{\pi /2}{\cos }^3\theta d\theta ={\int }_0^1(1-u^2)du={\left[u-\frac{1}{3}{u}^3\right]}_0^1=\frac{2}{3}.$$

Example 3 (trig substitution). ${\int }_0^1\sqrt{1-x^2}dx$. Let $x=\sin \theta$, $dx=\cos \theta d\theta$; limits: $\theta \in [0,\frac{\pi }{2}]$:

$${\int }_0^1\sqrt{1-x^2}dx={\int }_0^{\pi /2}{\cos }^2\theta d\theta =\frac{1}{2}{\int }_0^{\pi /2}(1+\cos 2\theta )d\theta =\frac{\pi }{4}.$$

Applications

• Computing areas and arc lengths of curves defined by circles, ellipses, and other conic sections.
• Evaluating integrals arising in Fourier analysis (orthogonality of $\sin$ and $\cos$).
• Solving integrals in physics involving oscillatory motion or circular geometry.

Trade-offs

• The even/odd strategy requires recognising the parity of both exponents before choosing a method; mixing up the cases leads to dead ends.
• After a trigonometric substitution, back-substituting from $\theta$ to $x$ requires care with signs (e.g.\ $\sqrt{1-{\sin }^2\theta }=|\cos \theta |$, not always $\cos \theta$).
• Trig substitution introduces a change of variable that must be invertible on the integration domain; checking the domain of $\theta$ is essential.
• For mixed products ${\sin }^m\theta {\cos }^n\theta$ with large even exponents, repeated application of half-angle formulas can become tedious; reduction formulas or a CAS may be preferable.

Links

• Integration by Substitution
• Integrals