Trigonometric Integrals

Cheatsheet Content

### Introduction to Trigonometric Integrals Trigonometric integrals are a class of integrals involving trigonometric functions. Solving them often requires strategic use of trigonometric identities, substitution techniques, and integration by parts. This cheatsheet covers common strategies for various forms of trigonometric integrals. ### Products of Powers of Sines and Cosines: $\int \sin^m x \cos^n x \, dx$ The strategy depends on the parity of $m$ and $n$. #### Case 1: At least one exponent is odd 1. **If $m$ is odd:** * Save one $\sin x$ factor. * Convert the remaining $\sin^m x$ factors to powers of $\cos x$ using $\sin^2 x = 1 - \cos^2 x$. * Let $u = \cos x$, then $du = -\sin x \, dx$. * Example: $\int \sin^3 x \cos^2 x \, dx$ * $\int \sin^2 x \cos^2 x \sin x \, dx = \int (1 - \cos^2 x) \cos^2 x \sin x \, dx$ * Let $u = \cos x$, $du = -\sin x \, dx$. * $\int (1 - u^2) u^2 (-du) = \int (u^4 - u^2) \, du = \frac{u^5}{5} - \frac{u^3}{3} + C$ * Substitute back: $\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$ 2. **If $n$ is odd:** * Save one $\cos x$ factor. * Convert the remaining $\cos^n x$ factors to powers of $\sin x$ using $\cos^2 x = 1 - \sin^2 x$. * Let $u = \sin x$, then $du = \cos x \, dx$. * Example: $\int \sin^4 x \cos^3 x \, dx$ * $\int \sin^4 x \cos^2 x \cos x \, dx = \int \sin^4 x (1 - \sin^2 x) \cos x \, dx$ * Let $u = \sin x$, $du = \cos x \, dx$. * $\int u^4 (1 - u^2) \, du = \int (u^4 - u^6) \, du = \frac{u^5}{5} - \frac{u^7}{7} + C$ * Substitute back: $\frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C$ #### Case 2: Both $m$ and $n$ are even * Use half-angle identities to reduce the powers: * $\sin^2 x = \frac{1 - \cos(2x)}{2}$ * $\cos^2 x = \frac{1 + \cos(2x)}{2}$ * $\sin x \cos x = \frac{\sin(2x)}{2}$ * Repeat until powers are manageable. * Example: $\int \sin^2 x \cos^2 x \, dx$ * $\int \left(\frac{1 - \cos(2x)}{2}\right) \left(\frac{1 + \cos(2x)}{2}\right) \, dx = \frac{1}{4} \int (1 - \cos^2(2x)) \, dx$ * $\frac{1}{4} \int \sin^2(2x) \, dx = \frac{1}{4} \int \frac{1 - \cos(4x)}{2} \, dx$ * $\frac{1}{8} \int (1 - \cos(4x)) \, dx = \frac{1}{8} \left(x - \frac{\sin(4x)}{4}\right) + C$ ### Eliminating Square Roots Square roots involving trigonometric expressions can often be simplified using identities. #### Form 1: $\sqrt{a^2 - x^2}$ * Let $x = a \sin \theta$. Then $dx = a \cos \theta \, d\theta$. * $\sqrt{a^2 - a^2 \sin^2 \theta} = \sqrt{a^2 (1 - \sin^2 \theta)} = \sqrt{a^2 \cos^2 \theta} = a |\cos \theta|$. * Assume $\theta \in [-\pi/2, \pi/2]$ so $\cos \theta \ge 0$, then $\sqrt{a^2 - x^2} = a \cos \theta$. #### Form 2: $\sqrt{a^2 + x^2}$ * Let $x = a \tan \theta$. Then $dx = a \sec^2 \theta \, d\theta$. * $\sqrt{a^2 + a^2 \tan^2 \theta} = \sqrt{a^2 (1 + \tan^2 \theta)} = \sqrt{a^2 \sec^2 \theta} = a |\sec \theta|$. * Assume $\theta \in (-\pi/2, \pi/2)$ so $\sec \theta \ge 0$, then $\sqrt{a^2 + x^2} = a \sec \theta$. #### Form 3: $\sqrt{x^2 - a^2}$ * Let $x = a \sec \theta$. Then $dx = a \sec \theta \tan \theta \, d\theta$. * $\sqrt{a^2 \sec^2 \theta - a^2} = \sqrt{a^2 (\sec^2 \theta - 1)} = \sqrt{a^2 \tan^2 \theta} = a |\tan \theta|$. * Assume $\theta \in [0, \pi/2)$ or $\theta \in [\pi, 3\pi/2)$ so $\tan \theta \ge 0$, then $\sqrt{x^2 - a^2} = a \tan \theta$. #### Example: $\int \frac{1}{\sqrt{9 - x^2}} \, dx$ * This is of the form $\sqrt{a^2 - x^2}$ with $a=3$. * Let $x = 3 \sin \theta$, $dx = 3 \cos \theta \, d\theta$. * $\sqrt{9 - x^2} = \sqrt{9 - 9 \sin^2 \theta} = \sqrt{9 \cos^2 \theta} = 3 \cos \theta$. * $\int \frac{3 \cos \theta}{3 \cos \theta} \, d\theta = \int 1 \, d\theta = \theta + C$. * Since $x = 3 \sin \theta$, then $\sin \theta = x/3$, so $\theta = \arcsin(x/3)$. * Result: $\arcsin(x/3) + C$. ### Integrals of Powers of $\tan x$ and $\sec x$: $\int \tan^m x \sec^n x \, dx$ The strategy depends on the parity of $m$ and $n$. #### Case 1: $n$ (power of $\sec x$) is even 1. Save a $\sec^2 x$ factor. 2. Convert the remaining $\sec^{n-2} x$ factors to powers of $\tan x$ using $\sec^2 x = 1 + \tan^2 x$. 3. Let $u = \tan x$, then $du = \sec^2 x \, dx$. 4. Example: $\int \tan^2 x \sec^4 x \, dx$ * $\int \tan^2 x \sec^2 x \sec^2 x \, dx = \int \tan^2 x (1 + \tan^2 x) \sec^2 x \, dx$ * Let $u = \tan x$, $du = \sec^2 x \, dx$. * $\int u^2 (1 + u^2) \, du = \int (u^2 + u^4) \, du = \frac{u^3}{3} + \frac{u^5}{5} + C$ * Substitute back: $\frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C$ #### Case 2: $m$ (power of $\tan x$) is odd 1. Save a $\sec x \tan x$ factor. 2. Convert the remaining $\tan^{m-1} x$ factors to powers of $\sec x$ using $\tan^2 x = \sec^2 x - 1$. 3. Let $u = \sec x$, then $du = \sec x \tan x \, dx$. 4. Example: $\int \tan^3 x \sec x \, dx$ * $\int \tan^2 x (\sec x \tan x) \, dx = \int (\sec^2 x - 1) (\sec x \tan x) \, dx$ * Let $u = \sec x$, $du = \sec x \tan x \, dx$. * $\int (u^2 - 1) \, du = \frac{u^3}{3} - u + C$ * Substitute back: $\frac{\sec^3 x}{3} - \sec x + C$ #### Case 3: $m$ is even and $n$ is odd (both non-zero) * This is generally the most difficult case. * Convert $\tan^m x$ to powers of $\sec x$ using $\tan^2 x = \sec^2 x - 1$. * This leaves you with powers of $\sec x$. Use reduction formulas for $\int \sec^n x \, dx$: * $\int \sec^n x \, dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} x \, dx$ (for $n \ge 2$) * Base cases: * $\int \sec x \, dx = \ln |\sec x + \tan x| + C$ * $\int \sec^2 x \, dx = \tan x + C$ * Example: $\int \tan^2 x \sec x \, dx$ * $\int (\sec^2 x - 1) \sec x \, dx = \int (\sec^3 x - \sec x) \, dx$ * $\int \sec^3 x \, dx - \int \sec x \, dx$ * Use reduction for $\sec^3 x$: $\frac{\sec x \tan x}{2} + \frac{1}{2} \int \sec x \, dx$ * So, $\left(\frac{\sec x \tan x}{2} + \frac{1}{2} \int \sec x \, dx\right) - \int \sec x \, dx$ * $\frac{\sec x \tan x}{2} - \frac{1}{2} \int \sec x \, dx = \frac{\sec x \tan x}{2} - \frac{1}{2} \ln |\sec x + \tan x| + C$ #### Case 4: Only powers of $\tan x$ ($\int \tan^m x \, dx$) * Save a $\tan^2 x$ factor. * Convert using $\tan^2 x = \sec^2 x - 1$. * Repeat, or use reduction formula: * $\int \tan^m x \, dx = \frac{\tan^{m-1} x}{m-1} - \int \tan^{m-2} x \, dx$ (for $m \ge 2$) * Base cases: * $\int \tan x \, dx = \ln |\sec x| + C$ or $-\ln |\cos x| + C$ * $\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x + C$ ### Products of Sines and Cosines (Different Arguments): $\int \sin(Ax)\cos(Bx) \, dx$, etc. Use product-to-sum identities: * $\sin A \cos B = \frac{1}{2}[\sin(A-B) + \sin(A+B)]$ * $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$ * $\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$ #### Example: $\int \sin(5x) \cos(3x) \, dx$ * Using $\sin A \cos B = \frac{1}{2}[\sin(A-B) + \sin(A+B)]$ with $A=5x, B=3x$: * $\int \frac{1}{2}[\sin(5x-3x) + \sin(5x+3x)] \, dx = \frac{1}{2} \int [\sin(2x) + \sin(8x)] \, dx$ * $\frac{1}{2} \left[-\frac{\cos(2x)}{2} - \frac{\cos(8x)}{8}\right] + C$ * $-\frac{\cos(2x)}{4} - \frac{\cos(8x)}{16} + C$ ### Trigonometric Substitutions Used for integrals containing expressions of the form $\sqrt{a^2 \pm x^2}$ or $\sqrt{x^2 - a^2}$. These were covered in "Eliminating Square Roots" but are crucial to reiterate as a standalone technique. #### Summary of Substitutions | Expression | Substitution | $dx$ | Identity Used | Resulting $\sqrt{\dots}$ | |--------------------|-------------------|-----------------------------|----------------------|--------------------------| | $\sqrt{a^2 - x^2}$ | $x = a \sin \theta$ | $dx = a \cos \theta \, d\theta$ | $1 - \sin^2 \theta = \cos^2 \theta$ | $a \cos \theta$ | | $\sqrt{a^2 + x^2}$ | $x = a \tan \theta$ | $dx = a \sec^2 \theta \, d\theta$ | $1 + \tan^2 \theta = \sec^2 \theta$ | $a \sec \theta$ | | $\sqrt{x^2 - a^2}$ | $x = a \sec \theta$ | $dx = a \sec \theta \tan \theta \, d\theta$ | $\sec^2 \theta - 1 = \tan^2 \theta$ | $a \tan \theta$ | #### Steps 1. Identify the form of the radical. 2. Choose the appropriate substitution for $x$ and $dx$. 3. Replace all $x$ terms in the integral with $\theta$ terms. Simplify the radical. 4. Evaluate the resulting trigonometric integral. 5. Substitute back to $x$ using a reference triangle or the original substitution. #### Example: $\int \frac{1}{(x^2+4)^{3/2}} \, dx$ * This is of the form $(a^2 + x^2)^{3/2} = (\sqrt{a^2 + x^2})^3$ with $a=2$. * Let $x = 2 \tan \theta$, $dx = 2 \sec^2 \theta \, d\theta$. * $x^2+4 = (2 \tan \theta)^2 + 4 = 4 \tan^2 \theta + 4 = 4(\tan^2 \theta + 1) = 4 \sec^2 \theta$. * $(x^2+4)^{3/2} = (4 \sec^2 \theta)^{3/2} = (\sqrt{4 \sec^2 \theta})^3 = (2 \sec \theta)^3 = 8 \sec^3 \theta$. * $\int \frac{2 \sec^2 \theta}{8 \sec^3 \theta} \, d\theta = \int \frac{1}{4 \sec \theta} \, d\theta = \frac{1}{4} \int \cos \theta \, d\theta$ * $\frac{1}{4} \sin \theta + C$. * From $x = 2 \tan \theta$, we have $\tan \theta = x/2$. Construct a right triangle: * Opposite $= x$, Adjacent $= 2$, Hypotenuse $= \sqrt{x^2+4}$. * So, $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{x}{\sqrt{x^2+4}}$. * Result: $\frac{1}{4} \frac{x}{\sqrt{x^2+4}} + C$. ### U-Substitution for Trig Integrals U-substitution is fundamental for simplifying many trigonometric integrals, often when an integrand contains a function and its derivative. #### General Form $\int f(g(x)) g'(x) \, dx$. Let $u = g(x)$, then $du = g'(x) \, dx$. This transforms the integral to $\int f(u) \, du$. #### Common Scenarios * **Inner function is linear:** $\int \cos(ax+b) \, dx$. Let $u = ax+b$, $du = a \, dx$. $\int \cos u \frac{1}{a} \, du = \frac{1}{a} \sin u + C = \frac{1}{a} \sin(ax+b) + C$. * **Function and its derivative present:** * $\int \sin^m x \cos x \, dx$. Let $u = \sin x$, $du = \cos x \, dx$. $\int u^m \, du = \frac{u^{m+1}}{m+1} + C$. * $\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$. Let $u = \cos x$, $du = -\sin x \, dx$. $\int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C$. * $\int \sec^n x \tan x \, dx$. Let $u = \sec x$, $du = \sec x \tan x \, dx$. $\int u^{n-1} \, du = \frac{u^n}{n} + C$. (If $n=1$, it's $\int \tan x \, dx$, special case). #### Example: $\int \frac{\sec^2 x}{\sqrt{\tan x}} \, dx$ * Let $u = \tan x$, $du = \sec^2 x \, dx$. * $\int \frac{1}{\sqrt{u}} \, du = \int u^{-1/2} \, du = 2u^{1/2} + C$ * Substitute back: $2\sqrt{\tan x} + C$. #### Example: $\int e^{\sin x} \cos x \, dx$ * Let $u = \sin x$, $du = \cos x \, dx$. * $\int e^u \, du = e^u + C$ * Substitute back: $e^{\sin x} + C$. ### Integration by Parts (IBP): $\int u \, dv = uv - \int v \, du$ IBP is crucial when the integrand is a product of two functions that don't fit a simple u-substitution pattern, or when integrating inverse trig functions or logarithms. #### Choosing $u$ and $dv$ (LIATE Rule) A mnemonic for prioritizing $u$: **L**ogarithmic, **I**nverse Trig, **A**lgebraic, **T**rigonometric, **E**xponential. Choose $u$ as the function that appears earliest in this list. #### Common Scenarios for Trig Integrals 1. **Inverse Trigonometric Functions:** When integrating $\arcsin x$, $\arctan x$, etc. * Example: $\int \arctan x \, dx$ * Let $u = \arctan x$, $dv = dx$. * Then $du = \frac{1}{1+x^2} \, dx$, $v = x$. * $x \arctan x - \int \frac{x}{1+x^2} \, dx$. * For the remaining integral, use u-sub: Let $w = 1+x^2$, $dw = 2x \, dx$. * $\int \frac{x}{1+x^2} \, dx = \int \frac{1}{w} \frac{dw}{2} = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+x^2) + C$. * Result: $x \arctan x - \frac{1}{2} \ln(1+x^2) + C$. 2. **Products of Polynomials and Trig Functions:** * Example: $\int x \cos x \, dx$ * Let $u = x$ (Algebraic), $dv = \cos x \, dx$ (Trigonometric). * Then $du = dx$, $v = \sin x$. * $x \sin x - \int \sin x \, dx = x \sin x - (-\cos x) + C = x \sin x + \cos x + C$. 3. **Cyclic Integrals:** Integrals that return to their original form after two applications of IBP. * Example: $\int e^x \sin x \, dx$ * Let $u = \sin x$, $dv = e^x \, dx$. Then $du = \cos x \, dx$, $v = e^x$. * $\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx$. * Apply IBP again to $\int e^x \cos x \, dx$: * Let $u = \cos x$, $dv = e^x \, dx$. Then $du = -\sin x \, dx$, $v = e^x$. * $\int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx$. * Substitute this back into the first equation: * $\int e^x \sin x \, dx = e^x \sin x - (e^x \cos x + \int e^x \sin x \, dx)$ * $\int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx$. * Let $I = \int e^x \sin x \, dx$. Then $I = e^x \sin x - e^x \cos x - I$. * $2I = e^x (\sin x - \cos x)$. * $I = \frac{e^x}{2} (\sin x - \cos x) + C$. (Don't forget the $+C$ at the end!) ### Summary of Strategies 1. **Simplify First:** Use algebraic manipulation or trigonometric identities before applying integration techniques. 2. **U-Substitution:** Look for a function and its derivative. This is often the first step to try. 3. **Trigonometric Identities:** Essential for simplifying products of powers of sines/cosines/tangents/secants, and for product-to-sum formulas. 4. **Trigonometric Substitution:** Used for integrals involving $\sqrt{a^2 \pm x^2}$ or $\sqrt{x^2 - a^2}$. 5. **Integration by Parts:** For products of unlike functions (e.g., $x \sin x$, $e^x \cos x$, $\ln x$, $\arctan x$). 6. **Reduction Formulas:** For higher powers of $\sin x, \cos x, \tan x, \sec x$. 7. **Completing the Square:** If you have an expression like $\sqrt{x^2+bx+c}$, complete the square to get it into one of the standard trig substitution forms. #### General Tips * **Know your identities:** $\sin^2 x + \cos^2 x = 1$, $\tan^2 x + 1 = \sec^2 x$, $\cot^2 x + 1 = \csc^2 x$, half-angle formulas, product-to-sum formulas. * **Practice:** Recognize patterns. The more you practice, the easier it becomes to identify the correct strategy. * **Don't forget the $+C$!** * **Reference Triangle:** Always draw a reference triangle when back-substituting after a trigonometric substitution. ### Common Trigonometric Derivatives and Integrals #### Derivatives * $\frac{d}{dx}(\sin x) = \cos x$ * $\frac{d}{dx}(\cos x) = -\sin x$ * $\frac{d}{dx}(\tan x) = \sec^2 x$ * $\frac{d}{dx}(\cot x) = -\csc^2 x$ * $\frac{d}{dx}(\sec x) = \sec x \tan x$ * $\frac{d}{dx}(\csc x) = -\csc x \cot x$ #### Integrals * $\int \sin x \, dx = -\cos x + C$ * $\int \cos x \, dx = \sin x + C$ * $\int \tan x \, dx = \ln |\sec x| + C$ or $-\ln |\cos x| + C$ * $\int \cot x \, dx = \ln |\sin x| + C$ or $-\ln |\csc x| + C$ * $\int \sec x \, dx = \ln |\sec x + \tan x| + C$ * $\int \csc x \, dx = \ln |\csc x - \cot x| + C$ * $\int \sec^2 x \, dx = \tan x + C$ * $\int \csc^2 x \, dx = -\cot x + C$ * $\int \sec x \tan x \, dx = \sec x + C$ * $\int \csc x \cot x \, dx = -\csc x + C$ #### General Derivative Rules * **Constant Rule:** $\frac{d}{dx}(c) = 0$ * **Power Rule:** $\frac{d}{dx}(x^n) = nx^{n-1}$ * **Constant Multiple Rule:** $\frac{d}{dx}(cf(x)) = c f'(x)$ * **Sum/Difference Rule:** $\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)$ * **Product Rule:** $\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$ * **Quotient Rule:** $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ * **Chain Rule:** $\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)$ * **Exponential Rule:** $\frac{d}{dx}(e^x) = e^x$ * **Logarithmic Rule:** $\frac{d}{dx}(\ln x) = \frac{1}{x}$ #### Inverse Trigonometric Derivatives * $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$ * $\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}$ * $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$ * $\frac{d}{dx}(\text{arccot } x) = -\frac{1}{1+x^2}$ * $\frac{d}{dx}(\text{arcsec } x) = \frac{1}{|x|\sqrt{x^2-1}}$ * $\frac{d}{dx}(\text{arccsc } x) = -\frac{1}{|x|\sqrt{x^2-1}}$ #### Inverse Trigonometric Integrals (Forms from Derivatives) * $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$ (for $a>0$) * $\int \frac{1}{a^2+x^2} \, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$ (for $a \ne 0$) * $\int \frac{1}{x\sqrt{x^2-a^2}} \, dx = \frac{1}{a}\text{arcsec}\left(\frac{|x|}{a}\right) + C$ (for $a>0$)