Lecture #8: Trigonometric Identities & Trigonometric Integrals

Right Triangle Trigonometry

For this definition, we assume \(0 \le \theta \le \frac{\pi}{2}\) or \(0^\circ \le \theta \le 90^\circ\).

Right Triangle Trigonometry

Right Triangle Trigonometry

For this definition, \(\theta\) can take any value.

Right Triangle Trigonometry

Trigonometric Identities

Tangent and Cotangent Identities

\[ \begin{aligned} \tan(\theta) &= \frac{\sin(\theta)}{\cos(\theta)} & \cot(\theta) &= \frac{\cos(\theta)}{\sin(\theta)} \end{aligned} \]

Reciprocal Identities

\[ \begin{aligned} \sin(\theta) &= \frac{1}{\csc(\theta)} & \csc(\theta) &= \frac{1}{\sin(\theta)} \\ \cos(\theta) &= \frac{1}{\sec(\theta)} & \sec(\theta) &= \frac{1}{\cos(\theta)} \\ \tan(\theta) &= \frac{1}{\cot(\theta)} & \cot(\theta) &= \frac{1}{\tan(\theta)} \end{aligned} \]

Trigonometric Identities

Pythagorean Identities

\[ \begin{aligned} \sin^2(\theta) + \cos^2(\theta) = 1 \\ \tan^2(\theta) + 1 = \sec^2(\theta) \\ 1 + \cot^2(\theta) = \csc^2(\theta) \end{aligned} \]

Even/Odd Formulas

\[ \begin{aligned} \sin(-\theta) &= -\sin(\theta) & \csc(-\theta) &= -\csc(\theta) \\ \cos(-\theta) &= \cos(\theta) & \sec(-\theta) &= \sec(\theta) \\ \tan(-\theta) &= -\tan(\theta) & \cot(-\theta) &= -\cot(\theta) \end{aligned} \]

Trigonometric Identities

Periodic Formulas} If \(n\) is an integer then:

\[ \begin{aligned} \sin(\theta + 2 \pi n) &= \sin(\theta) & \csc(\theta + 2 \pi n) = \csc(\theta) \\ \cos(\theta + 2 \pi n) &= \cos(\theta) & \sec(\theta + 2 \pi n) = \sec(\theta) \\ \tan(\theta + \pi n) &= \tan(\theta) & \cot(\theta + \pi n) = \cot(\theta) \end{aligned} \]

Sum and Difference Formulas

\[ \begin{aligned} \sin(\alpha \pm \beta) &= \sin(\alpha)\cos(\beta) \pm \cos(\alpha)\sin(\beta) \\ \cos(\alpha \pm \beta) &= \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta) \\ \tan(\alpha \pm \beta) &= \frac{\tan(\alpha) \pm \tan(\beta)}{1 \mp \tan(\alpha)\tan(\beta)} \end{aligned} \]

Trigonometric Identities

Half Angle Formulas

\[ \begin{aligned} \begin{array}{r} \sin\left( \dfrac{\theta}{2} \right) = \pm \sqrt{\dfrac{1 - \cos(\theta)}{2}} \\[1em] \cos\left( \dfrac{\theta}{2} \right) = \pm \sqrt{\dfrac{1 + \cos(\theta)}{2}} \end{array} \quad \Longrightarrow \quad \tan\left( \frac{\theta}{2} \right) &= \pm \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \end{aligned} \]

Trigonometric Identities

Half Angle Formulas (alternate form)

\[ \begin{aligned} \begin{array}{r} \sin^2(\theta) = \dfrac{1 - cos(2 \theta)}{2} \\[1em] \cos^2(\theta) = \dfrac{1 + cos(2 \theta)}{2} \end{array} \quad \Longrightarrow \quad \tan^2(\theta) = \dfrac{1 - cos(2\theta)}{1 + \cos(2 \theta)} \end{aligned} \]

Calculus on Trigonometric Functions

We know some basic formulas for trigonometric derivatives and antiderivatives:

\[ \begin{aligned} &\frac{d}{dx} \, \sin(x) = \cos(x) & &\int \cos(x) \, dx = \sin(x) + C \\ &\frac{d}{dx} \, \cos(x) = -\sin(x) & &\int \sin(x) \, dx = -\cos(x) + C \\ &\frac{d}{dx} \, \tan(x) = \sec^2(x) & &\int \sec^2(x) \, dx = \tan(x) + C \\ \end{aligned} \]

Calculus on Trigonometric Functions

We know some basic formulas for trigonometric derivatives and antiderivatives:

\[ \begin{aligned} &\frac{d}{dx} \, \sec(x) = \sec(x) \tan(x) & &\int \sec(x) \tan(x) \, dx = \sec(x) + C \\ &\frac{d}{dx} \, \csc(x) = -\csc(x)\cot(x) & &\int \csc(x)\cot(x) \, dx = -\csc(x) + C \\ &\frac{d}{dx} \, \cot(x) = -\csc^2(x) & &\int \csc^2(x) \, dx = -\cot(x) + C \\ \end{aligned} \]

Powers of Sines and Cosines

To integrate \(\displaystyle\int \cos^j(x) \sin^k(x) \, dx\), use the following strategies:

1. If \(k\) is odd, rewrite \(\sin^k(x)\) as \(\sin^{k-1}(x) \sin(x)\) and use the identity \(\sin^2(x) = 1 - \cos^2(x)\) to rewrite \(\sin^{k-1}(x)\) in terms of \(\cos(x)\). Integrate using the substitution \(u = \cos(x)\). This substitution makes \(du = -\sin(x) \, dx\).

Powers of Sines and Cosines

To integrate \(\displaystyle\int \cos^j(x) \sin^k(x) \, dx\), use the following strategies:

2. If \(j\) is odd, rewrite \(\cos^j(x)\) as \(\cos^{j-1}(x) \cos(x)\) and use the identity \(\cos^2(x) = 1 - \sin^2(x)\) to rewrite \(\cos^{j-1}(x)\) in terms of \(\sin(x)\). Integrate using the substitution \(u = \sin(x)\). This substitution makes \(du = \cos(x) \, dx\). (Note: If both \(j\) and \(k\) are odd, either strategy 1 or strategy 2 may be used.)

Powers of Sines and Cosines

To integrate \(\displaystyle\int \cos^j(x) \sin^k(x) \, dx\), use the following strategies:

3. If both \(j\) and \(k\) are even, use \(\sin^2(x) = \dfrac{1}{2} - \dfrac{1}{2}\cos(2x)\) and \(\cos^2(x) = \dfrac{1}{2} + \dfrac{1}{2} \cos(2x)\). After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.

Powers of Sines and Cosines

Example: Evaluate \(\displaystyle\int \cos^3(x)\sin^2(x) \, dx\).

Products of \(\sin(ax)\) and \(\cos(bx)\)

\[ \begin{aligned} \sin(ax) \sin(bx) &= \frac{1}{2} \, \cos\big((a - b) \, x \big) - \frac{1}{2} \,\cos\big( (a + b) \, x \big) \\[2em] \sin(ax) \cos(bx) &= \frac{1}{2} \, \sin\big((a - b) \, x \big) + \frac{1}{2} \, \sin\big( (a + b) \, x \big) \\[2em] \cos(ax) \cos(bx) &= \frac{1}{2} \, \cos\big((a - b) \, x \big) + \frac{1}{2} \, \cos\big( (a + b) \, x \big) \end{aligned} \]

Powers of Sines and Cosines

Example: Evaluate \(\displaystyle\int \sin^2(x)\cos^2(x) \, dx\).

Powers of Tangents and Secants

\[ \begin{aligned} &\int \sec^2(x) \, dx = \tan(x) + C \\ &\int \sec(x) \tan(x) \, dx = \sec(x) + C \\ &\int \tan(x) \, dx = \ln \lvert \sec(x) \rvert + C \\ &\int \sec(x) \, dx = \ln \lvert \sec(x) + \tan(x) \rvert + C \end{aligned} \]

Integrals involving \(\displaystyle\int \tan^j(x) \sec(x)^2 \, dx\) and \(\displaystyle\int \sec^j(x) \tan(x) \, dx\) can also be done by substitution.

Powers of Tangents and Secants

To integrate \(\displaystyle\int \tan^k(x) \sec^j(x) \, dx\), use the following strategies:

1. If \(j\) is even and \(j \ge 2\), rewrite \(\sec^j(x)\) as \(\sec^{j-2}(x) \sec^2(x)\) and use \(\sec^2(x) = \tan^2(x) + 1\) to rewrite \(\sec^{j-2}(x)\) in terms of \(\tan(x)\). Let \(u = \tan(x)\) and \(du = \sec^2(x) \, dx\).

Powers of Tangents and Secants

To integrate \(\displaystyle\int \tan^k(x) \sec^j(x) \, dx\), use the following strategies:

2. If \(k\) is odd and \(j \ge 1\), rewrite \(\tan^k(x) \sec^j(x)\) as \(\tan^{k-1}(x) \sec^{j-1}(x) \sec(x) \tan(x)\) and use \(\tan^2(x) = \sec^2(x) - 1\) to rewrite \(\tan^{k-1}(x)\) in terms of \(\sec(x)\). Let \(u = \sec(x)\) and \(du = \sec(x)\tan(x) \, dx\). (Note: If \(j\) is even and \(k\) is odd, then either strategy 1 or strategy 2 may be used.)

Powers of Tangents and Secants

To integrate \(\displaystyle\int \tan^k(x) \sec^j(x) \, dx\), use the following strategies:

3. If \(k\) is odd where \(k \ge 3\) and \(j = 0\), rewrite \(\tan^k(x)\) as \(\tan^{k-2}(x)\sec^2(x) - \tan^{k-2}(x)\). It may be necessary to repeat this process on the \(\tan^{k-2}(x)\) term.

\[ \begin{aligned} \tan^k(x) &= \tan^{k-2}(x) \tan^2(x) \\ &= \tan^{k-2}(x) (\sec^2(x) - 1) \\ &= \tan^{k-2}(x)\sec^2(x) - \tan^{k-2}(x) \end{aligned} \]

Powers of Tangents and Secants

To integrate \(\displaystyle\int \tan^k(x) \sec^j(x) \, dx\), use the following strategies:

4. If \(k\) is even and \(j\) is odd, then use \(\tan^2(x)\) as \(\sec^2(x) - 1\) to express \(\tan^k(x)\) in terms of \(\sec(x)\). Use integration by parts to integrate odd powers of \(\sec(x)\).

Powers of Tangents and Secants

Example: Evaluate \(\displaystyle\int \tan^4(x)\sec^4(x) \, dx\).

Powers of Tangents and Secants

Using integration by parts (as before), we can derive the following for powers of secants or tangents.

\[ \begin{aligned} \int \sec^n(x) \, dx &= \frac{1}{n-1} \, \sec^{n-2}(x) \tan(x) + \frac{n-2}{n-1} \int \sec^{n-2}(x) \, dx \\[1em] \int \tan^n(x) \, dx &= \frac{1}{n-1} \tan^{n-1}(x) - \int \tan^{n-2}(x) \, dx \end{aligned} \]