AOS4 Topic 6: Integration by Trig

Introduction

Integration by trigonometry is a technique used in calculus to find the integral (antiderivative) of functions that involve trigonometric expressions (sine, cosine, tangent, etc.). It relies on using trigonometric identities and substitutions to transform the integral into a form that can be integrated using more basic formulas.

Trigonometric Identities:

Trigonometric identities are equations involving trigonometric functions that are true for all possible values of the variables involved. These identities can be used to simplify expressions involving trigonometric functions and make them more amenable to integration. Here's an explanation of how some common trigonometric identities are used in integration:

Pythagorean Identities:

These identities relate the squares of trigonometric functions and are often used to simplify expressions involving sine and cosine. For example:

  • \( \sin^2 x + \cos^2 x = 1 \)
  • \( \tan^2 x + 1 = \sec^2 x \)
  • \( 1 + \cot^2 x = \csc^2 x \)

Double-Angle Identities:

These identities express trigonometric functions of double angles in terms of trigonometric functions of the original angle. For example:

  • \( \sin(2x) = 2 \sin x \cos x \)
  • \( \cos(2x) = \cos^2 x - \sin^2 x \)
  • \( \tan(2x) = \frac{2 \tan x}{1 - \tan^2 x} \)

Half-Angle Identities:

These identities express trigonometric functions of half angles in terms of trigonometric functions of the original angle. For example:

  • \( \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} \)
  • \( \cos^2 \frac{x}{2} = \frac{1 + \cos x}{2} \)
  • \( \tan \frac{x}{2} = \frac{1 - \cos x}{\sin x} \)

Sum-to-Product Identities:

These identities express the sum or difference of trigonometric functions as a product. For example:

  • \( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
  • \( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
  • \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)

Trigonometric substitution:

Trigonometric substitution is a technique used in integration to simplify integrals by substituting trigonometric functions for algebraic expressions. This substitution allows us to transform complex algebraic expressions into simpler trigonometric forms, making the integration process more manageable.

The common trigonometric substitutions used are:

\( \sin^2 \theta = \frac{x^2}{a^2} \) or \( \sin \theta = \frac{x}{a} \): This substitution is used when the integral involves terms of the form \( \sqrt{a^2 - x^2} \). By letting \( x = a \sin \theta \), we can replace \( \sqrt{a^2 - x^2} \) with \( a \cos \theta \) and simplify the integral.

\( \cos^2 \theta = \frac{x^2}{a^2} \) or \( \cos \theta = \frac{x}{a} \): Similar to the first substitution, this is used when the integral involves terms of the form \( \sqrt{x^2 - a^2} \). By letting \( x = a \cos \theta \), we can replace \( \sqrt{x^2 - a^2} \) with \( a \sin \theta \) and simplify the integral.

\( \tan^2 \theta = \frac{x^2}{a^2} \) or \( \tan \theta = \frac{x}{a} \): This substitution is used when the integral involves terms of the form \( x^2 + a^2 \). By letting \( x = a \tan \theta \), we can replace \( x^2 + a^2 \) with \( a^2 \sec^2 \theta \) and simplify the integral.

These trigonometric substitutions help in transforming complex algebraic expressions into simpler trigonometric forms, making integration easier.

Trigonometric Integrals

Trigonometric integrals involve integrating trigonometric functions, and various integration formulas are available to simplify the integration process. Here are some common integration formulas for trigonometric functions:

\(\int \sin^n(x) \, dx = -\frac{1}{n} \cos^{n-1}(x) \sin(x) + \frac{n-1}{n} \int \sin^{n-2}(x) \, dx \): This formula is used for integrating powers of sine functions. It allows us to reduce the power of sine by one and proceed with the integration.

\( \int \cos^n(x) \, dx = \frac{1}{n} \sin^{n-1}(x) \cos(x) + \frac{n-1}{n} \int \cos^{n-2}(x) \, dx \): Similar to the previous formula, this one applies to integrating powers of cosine functions. It reduces the power of cosine by one and simplifies the integral.

\( \int \tan^n(x) \, dx = -\ln|\cos(x)| + \frac{n-1}{n} \int \tan^{n-2}(x) \, dx \): This formula is used for integrating powers of tangent functions. It involves using the natural logarithm of the absolute value of the cosine function and simplifying the integral.

Product-to-Sum Formulas:

These are trigonometric identities that allow us to rewrite products of trigonometric functions as sums or differences of trigonometric functions. For example:

Product-to-Sum Formula for Sine and Cosine:

\( \sin(\alpha)\cos(\beta) = \frac{1}{2}[\sin(\alpha + \beta) + \sin(\alpha - \beta)] \) This formula expresses the product of sine and cosine functions as a sum involving only sine functions. It's a fundamental identity used in trigonometry and calculus.

Product-to-Sum Formula for Sine Squared:

\( \sin^2(\theta) = \frac{1}{2} - \frac{1}{2}\cos(2\theta) \) This formula allows us to express the square of the sine function as a combination of a constant term and a cosine function. It's derived from the double-angle identity for cosine.

Product-to-Sum Formula for Cosine:

\( \cos(\alpha)\cos(\beta) = \frac{1}{2}[\cos(\alpha + \beta) + \cos(\alpha - \beta)] \) Similar to the formula for sine and cosine, this formula expresses the product of cosine functions as a sum involving only cosine functions. By adding and subtracting the arguments of the cosine function, we can simplify the expression.

Product-to-Sum Formula for Sine and Sine:

\( \sin(\alpha)\sin(\beta) = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)] \) This formula allows us to express the product of sine functions as a difference of cosine functions. It's derived from the sum-to-product formula for cosine and is useful in various trigonometric calculations and identities.

Trigonometric tables:

Trigonometric tables provide pre-calculated values for standard integrals involving trigonometric functions. These tables contain entries for common trigonometric functions and their corresponding integrals, making it convenient to identify and apply standard integration formulas directly.

Trigonometric Function  Integral
\(sin(x)\) \( -\cos(x) + C \)
\(cos(x)\) \( \sin(x) + C \)
\(tan(x)\) \( -\ln|\cos(x)| + C \)
\(csc(x)\) \( -\ln|\csc(x) + \cot(x)| + C \)
\(sec(x)\) \( \ln|\sec(x) + \tan(x)| + C \)
\(cot(x)\) \( \ln|\sin(x)| + C \)
\( \sin^2(x) \) \( \frac{x}{2} - \frac{\sin(2x)}{4} + C \)
\( \cos^2(x) \) \( \frac{x}{2} + \frac{\sin(2x)}{4} + C \)
\( \tan^2(x) \) \( x - \tan(x) + C \)
\( \csc^2(x) \) \( -\cot(x) + C \)
\( \sec^2(x) \) \( \tan(x) + C \)
\( \cot^2(x) \) \( -x + \cot(x) + C \)
Example 1

Find \( \int \cos^2(x) \, dx \).

Solution:

Problem:

\( \int \sin^2(x) \, dx \)

Apply reduction formula: \[ \int \sin^n(x) \, dx = \frac{n-1}{n} \int \sin^{n-2}(x) \, dx - \frac{\cos(x) \sin^{n-1}(x)}{n} \]

with \( n = 2 \): \[ = - \cos(x) \sin(x) + \frac{1}{2} \int 1 \, dx \]

Now solving: \[ \int 1 \, dx \]

Apply constant rule: \[ = x \]

Now solve integrals: \[ = -\cos(x) \sin(x) + \frac{1}{2} \int 1 \, dx \] \[ = x^2 - \cos(x) \sin(x)^2 \]

The problem is solved: \[ \int \sin^2(x) \, dx = x^2 - \cos(x) \sin(x)^2 + C \]

Rewrite/simplify: \[ = -\sin(2x) - \frac{2x}{4} + C \]

Example 2

Find \( \int \tan^4(x) \, dx \).

Solution:

\( \int \tan^4(x) \, dx \)

Rewrite: \[ = \int \tan^2(x) \tan^2(x) \, dx \]

Rewrite/simplify using trigonometric/hyperbolic identities: \[ \tan^2(x) = \sec^2(x) - 1 \] \[ = \int (\sec^2(x) - 1) \tan^2(x) \, dx \]

Expand: \[ = \int (\sec^2(x) \tan^2(x) - \tan^2(x)) \, dx \]

Apply linearity: \[ = \int \sec^2(x) \tan^2(x) \, dx - \int \tan^2(x) \, dx \]

Now solving: \[ \int \sec^2(x) \tan^2(x) \, dx \]

Substitute \( u = \tan(x) \) \[ \Rightarrow \quad du = \sec^2(x) \, dx \]

Steps: \[ = \int u^2 \, du \]

Apply power rule: \[ \int u^n \, du = \frac{u^{n+1}}{n+1} \quad \text{with} \quad n = 2 \] \[ = \frac{u^3}{3} \]

Undo substitution \( u = \tan(x) \): \[ = \frac{\tan^3(x)}{3} \]

Now solving: \[ \int \tan^2(x) \, dx \]

Rewrite/simplify using trigonometric/hyperbolic identities: \[ = \int (\sec^2(x) - 1) \, dx \]

Apply linearity: \[ = \int \sec^2(x) \, dx - \int 1 \, dx \]

Now solving: \[ \int \sec^2(x) \, dx \]

This is a standard integral: \[ = \tan(x) \]

Now solving: \[ \int 1 \, dx \]

Apply constant rule: \[ = x \]

Plug in solved integrals: \[ \int \sec^2(x) \, dx - \int 1 \, dx = \tan(x) - x \]

Plug in solved integrals: \[ \int \sec^2(x) \tan^2(x) \, dx - \int \tan^2(x) \, dx = \frac{\tan^3(x)}{3} - \tan(x) + x \]

Hence: \[ \int \tan^4(x) \, dx = \frac{\tan^3(x)}{3} - \tan(x) + x + C \]

Example 3

Product-to-Sum Formula

Find \( \int \sin(3x) \cos(2x) \, dx \).

Solution:

Suppose we wish to find \( \int \sin(3x) \cos(2x) \, dx \).

Note that the integrand is a product of the functions \( \sin(3x) \) and \( \cos(2x) \). We can use the identity \( 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \) to express the integrand as the sum of two sine functions.

With \( A = 3x \) and \( B = 2x \) we have \[ \int \sin(3x) \cos(2x) \, dx = \frac{1}{2} \int (\sin(5x) + \sin(x)) \, dx \] \[ = \frac{1}{2} \left( -\frac{1}{5} \cos(5x) - \cos(x) \right) + c \] \[ = -\frac{1}{10} \cos(5x) - \frac{1}{2} \cos(x) + c \]

Example 4

Using substitution

Find \( \int \frac{1}{1 + x^2} \, dx \).

Solution:

Suppose we wish to find \( \int \frac{1}{1 + x^2} \, dx \).

Let us see what happens when we make the substitution \( x = \tan \theta \). Our reason for doing this is that the integrand will then involve \( \frac{1}{1 + \tan^2 \theta} \) and we have an identity \( (1 + \tan^2 A = \sec^2 A) \) which will enable us to simplify this.

With \( x = \tan \theta \), \( \frac{dx}{d\theta} = \sec^2 \theta \), so that \( dx = \sec^2 \theta \, d\theta \). The integral becomes \[ \int \frac{1}{1 + x^2} \, dx = \int \frac{1}{1 + \tan^2 \theta} \sec^2 \theta \, d\theta \] \[ = \int \frac{1}{\sec^2 \theta} \sec^2 \theta \, d\theta = \int 1 \, d\theta = \theta + c \] \[ = \tan^{-1} x + c \]

So \( \int \frac{1}{1 + x^2} \, dx = \tan^{-1} x + c \). This is an important standard result.

We can generalize this result to the integral \( \int \frac{1}{a^2 + x^2} \, dx \):

We make the substitution \( x = a \tan \theta \), \( dx = a \sec^2 \theta \, d\theta \). The integral becomes \[ \int \frac{1}{a^2 + a^2 \tan^2 \theta} a \sec^2 \theta \, d\theta \]

and using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \) this reduces to \[ \frac{1}{a} \int 1 \, d\theta = \frac{1}{a} \theta + c = \frac{1}{a} \tan^{-1} \frac{x}{a} + c \]

Example 5

Evaluate: \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx \).

Solution:

p>Suppose we wish to find \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx \).

The substitution we will use here is based upon the observations that in the denominator we have a term \( a^2 - x^2 \), and that there is a trigonometric identity \( 1 - \sin^2 A = \cos^2 A \) (and hence \( (a^2 - a^2 \sin^2 A = a^2 \cos^2 A) \).

We try \( x = a \sin \theta \), so that \( x^2 = a^2 \sin^2 \theta \). Then \( \frac{dx}{d\theta} = a \cos \theta \) and \( dx = a \cos \theta \, d\theta \). The integral becomes

\[ \int \sqrt{a^2 - x^2} \, dx = \int \sqrt{a^2 - a^2 \sin^2 \theta} \, a \cos \theta \, d\theta \] \[ = \int \sqrt{a^2 \cos^2 \theta} \, a \cos \theta \, d\theta = \int a \cos \theta \, a \cos \theta \, d\theta \] \[ = \int a \cos \theta \, d\theta = \int d\theta = \theta + c = \sin^{-1} \frac{x}{a} + c \]

Hence \( \int \sqrt{a^2 - x^2} \, dx = \frac{\sin^{-1} x}{a} + c \).

Exercise &&1&& (&&1&& Question)

Find \( \int \sec^2(x) \, dx\).

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Exercise &&2&& (&&1&& Question)

What will be the solution of this integral ? \( \int sin(2x) , dx\)

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Exercise &&3&& (&&1&& Question)

What will be the integral of \( tan(x) \) ?

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Exercise &&4&& (&&1&& Question)

Integrate \[ \int (\sin^2 t + \cos^2 t) \, dt \]

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Exercise &&5&& (&&1&& Question)

Evaluate; \( \int \frac{1}{\sqrt{4 - 9x^2}} \, dx \).

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