Trigonometric Integrals and Substitution

The Pythagorean identity ${\sin }^{2}x+{\cos }^{2}x=1$ and its relatives let sine and cosine trade places, and trade with tangent and secant. Two techniques exploit this. The first evaluates integrals that are already trigonometric by rewriting the integrand until a substitution appears. The second introduces a trigonometric variable to remove a radical from an algebraic integrand.

Powers of sine and cosine

For $\int {\sin }^{m}x{\cos }^{n}x\mathbf{d}x$, use that the derivative of $\sin x$ is $\cos x$ and vice versa: one saved factor becomes the $\mathbf{d}u$ of a substitution while the rest converts to the other function.

One odd power

If the power of cosine is odd, split off a single $\cos x$ and convert the remaining even power with ${\cos }^{2}x=1-{\sin }^{2}x$; then substitute $u=\sin x$. For $\int {\cos }^{3}x\mathbf{d}x$,

With $u=\sin x$, $\mathbf{d}u=\cos x\mathbf{d}x$,

The same move handles a mix. For $\int {\sin }^{5}x{\cos }^{2}x\mathbf{d}x$ the sine power is odd, so save one $\sin x$ and convert ${\sin }^{4}x=(1-{\cos }^{2}x{)}^2$; substitute $u=\cos x$, $\mathbf{d}u=-\sin x\mathbf{d}x$:

Both powers even

When both powers are even, no single saved factor helps. Drop the exponent with the half-angle identities

For $\int {\sin }^{2}x\mathbf{d}x=\frac{1}{2}\int (1-\cos 2x)\mathbf{d}x=\frac{1}{2}(x-\frac{1}{2}\sin 2x)+C$. A fourth power needs the identity twice. Writing ${\sin }^{4}x=({\sin }^{2}x{)}^2$,

where the leftover ${\cos }^{2}2x$ was reduced by a second half-angle step.

Powers of tangent and secant

The same tactic works for $\int {\tan }^{m}x{\sec }^{n}x\mathbf{d}x$, using $\frac{\mathbf{d}}{\mathbf{d}x}\tan x={\sec }^{2}x$ and $\frac{\mathbf{d}}{\mathbf{d}x}\sec x=\sec x\tan x$ and the identity ${\sec }^{2}x=1+{\tan }^{2}x$.

• Even power of secant. Save ${\sec }^{2}x$, convert the rest to tangent, and substitute $u=\tan x$. For $\int {\tan }^{6}x{\sec }^{4}x\mathbf{d}x$, $$\int {\tan }^{6}x(1+{\tan }^{2}x){\sec }^{2}x\mathbf{d}x=\int (u^6+u^8)\mathbf{d}u=\frac{1}{7}{\tan }^{7}x+\frac{1}{9}{\tan }^{9}x+C.$$
• Odd power of tangent. Save $\sec x\tan x$, convert the remaining even power of tangent with ${\tan }^{2}x={\sec }^{2}x-1$, and substitute $u=\sec x$.

Two standard results are needed as building blocks. Integrating $\tan x$ gives a logarithm, and secant is handled by a multiply-and-substitute trick:

The secant formula follows by multiplying the integrand by $(\sec x+\tan x)/(\sec x+\tan x)$: the numerator becomes exactly the derivative of the denominator. When an odd power of secant appears alone, integration by parts produces a cyclic integral. For $\int {\sec }^{3}x\mathbf{d}x$, take $u=\sec x$, $\mathbf{d}v={\sec }^{2}x\mathbf{d}x$:

which solves to

Products of different frequencies

For $\int \sin mx\cos nx\mathbf{d}x$ and its cousins, the product-to-sum identities turn a product into a sum of single trigonometric terms:

with matching identities for $\sin A\sin B$ and $\cos A\cos B$. For $\int \sin 4x\cos 5x\mathbf{d}x=\frac{1}{2}\int (-\sin x+\sin 9x)\mathbf{d}x=\frac{1}{2}(\cos x-\frac{1}{9}\cos 9x)+C$.

Trigonometric substitution

Radicals of the form $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, and $\sqrt{x^2-a^2}$ block both substitution and parts. Replacing $x$ by a trigonometric function of a new variable clears the root through a Pythagorean identity. The substitution runs backward from the usual direction: the old variable $x$ becomes a function of the new one, called an inverse substitution.

For $\sqrt{a^2-x^2}$, set $x=a\sin \theta$ with $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$. Then

with $\cos \theta \ge 0$ on that interval, so the radical disappears. The restriction on $\theta$ makes the substitution one-to-one, matching the range of the inverse function needed to return to $x$.

Each substitution corresponds to a right triangle whose sides encode the relation between $x$, $a$, and the radical. After integrating in $\theta$, the triangle converts $\theta$ and every trigonometric function of it back to $x$.

A sine substitution

The substitution mirrors the geometry of a circle: $x=a\sin \theta$ makes $x$ the side opposite $\theta$, $a$ the hypotenuse (the radius), and $\sqrt{a^2-x^2}$ the adjacent side.

A tangent substitution

A denominator with $\sqrt{x^2+a^2}$ calls for $x=a\tan \theta$.

A secant substitution

For $\int \frac{\mathbf{d}x}{\sqrt{x^2-a^2}}$ with $a>0$, set $x=a\sec \theta$, $\mathbf{d}x=a\sec \theta \tan \theta \mathbf{d}\theta$, and $\sqrt{x^2-a^2}=a\tan \theta$:

The triangle gives $\sec \theta =x/a$ and $\tan \theta =\sqrt{x^2-a^2}/a$, so

after absorbing the constant $-\ln a$ into $C$. Not every quadratic radical needs trigonometry: $\int x/\sqrt{x^2+4}\mathbf{d}x$ has the extra $x$ that makes the plain substitution $u=x^2+4$ simpler. Reach for a trigonometric substitution only when the direct one fails.

Completing the square

A radical like $\sqrt{3-2x-x^2}$ is not yet in standard form, but completing the square turns it into one:

The substitution $u=x+1$ reduces $\sqrt{4-(x+1{)}^2}$ to $\sqrt{4-u^2}$, which then takes $u=2\sin \theta$. Completing the square routes any $\sqrt{a{x}^2+bx+c}$ into one of the three standard radicals.

The area of an ellipse

Trigonometric substitution recovers the area of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Solving for $y$ in the first quadrant gives $y=\frac{b}{a}\sqrt{a^2-x^2}$, and by symmetry the total area is four times the first-quadrant piece:

Substituting $x=a\sin \theta$ (so $\mathbf{d}x=a\cos \theta \mathbf{d}\theta$) and changing the limits to $\theta =0$ and $\theta =\frac{\pi }{2}$,

Setting $a=b=r$ gives the area of a circle, $\pi r^2$.

Which technique

Trigonometric integrals and trigonometric substitution run in opposite directions. The first removes trigonometric functions by converting an integrand to a polynomial in $\sin x$ or $\tan x$; the second introduces trigonometric functions to remove a radical. Both use the same three Pythagorean identities, read in whichever direction the integrand requires. Ratios of polynomials, the remaining large class of elementary integrands, yield to the partial-fraction method.¹²