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Integrating trigonometric expressions 1: Concise Easy

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Integrating trigonometric expressions involves solving trigonometric expressions that contain trigonometric functions. These functions are periodical. They denote the relationship between the angle and the sides of a right-angled triangle.

Trigonometric rations are used to express sides of a right angled triangle

Examples of trigonometric functions includes:

  • sine
  • cosines
  • tangents
  • secants
  • cosecants
  • cotangents

There are various techniques used to integrate mathematical expressions with trigonometric functions. Integration of trigonometric expressions requires a mix of direct integration rules. This may includes simplification using identities, substitution, and sometimes integration by parts, depending on the form of the expression. some of the techniques includes:

1.Memorizing and quoting some know trigonometric functions and identities

some trigonometric identities are known to transform to another specific and definite trigonometric functions after the integration process. This can be very useful during Integrating trigonometric expressions. Some examples includes:

$$\text{(i).} \int sin x dx = -cos x +c $$

The above expression shows that we always obtain a negative value of a cosine function. This negative value is accompanied by another constant number whenever the sine function is integrated.

$$\text{(ii).} \int cos x dx = sinx +c $$

in other words, integrating a cosine functions results to a positive value of a cosine function when Integrating trigonometric expressions.

$$\text{(iii).} \int sec^2 x dx = tan x +c $$
$$\text{(iv).} \int csc^2 x dx = -cot x +c $$
$$\text{(v).} \int sec x tanx dx = -secx +c $$
$$\text{(vi).} \int csc x cotxdx = -cscx +c $$

2. Using trigonometric identities

by using trigonometric identities, some trigonometric expressions that could be complex to integrate becomes simplified .

When Integrating trigonometric expressions useful trigonometric identities includes:

$$\text{(i.)} \ sin^2x = \frac{1-cos 2x}{2} $$ $$\ cos^2x = \frac{1+cos 2x}{2}$$
$$\text{(ii.)}sin x cosx = \frac{1}{2}sin2x$$

The above identity is known as power reduction identity

3. Substitution Method

Substitution method in integration is a used to simplify an integral by making a change of variables. This transforms the integral into a simpler form Integrating trigonometric expressions. It is often used when the integrand contains a composite function, making it easier to integrate.

Substitution generally has the following steps:

  • Identifying part of the integrand that can be replaced with a new variable (usually denoted by u).
  • Compute the derivative of the chosen substitution with respect to x (i.e., find du). This helps in expressing dx in terms of du.
  • Substitute u and du into the original integral, replacing the x-terms with the new u-terms. This aims at simplifying the integral so that it’s easier to solve.
  • Perform the integration with respect to u.
  • Once the integral is solved in terms of u, substitute the original expression for u back into the result to return to the variable x.

In trigonometric functions, the trigonometric ratio can have even or odd powers, which determines how they are treated during integration.

4. Integration by Parts

Integration by parts is a technique commonly used to integrate product of two functions. It is based on the product rule for differentiation derived from the following formula:

$$\int \ udv = uv – \int v \ du $$

By applying the formula the above formula, you can reduce the complexity of the integral so that it is easier to solve.

Integration by parts may involves steps like:

  • Choose two functions from the integrand, u and dv, such that the integral of v is easier to compute than the original one
  • Compute du, the derivative of u.
  • Compute v, the integral of dv
  • Substitute u, dv, du, and v into the integration by parts formula.
  • If the resulting integral is easier to evaluate, solve it. If not, apply integration by parts again

5. Trigonometric Substitution

This is used when the integrand involves square roots and trigonometric functions. consider the integral:

$$\int \frac{dx}{\sqrt{1-x^2}}$$

An integral that looks like the above expression can be solved first by substituting x for sinθ such that x -sinθ = 0

Integrating Even powers of sine and cosine

Integrating even powers of sine and cosine involves simplifying the integral of expressions where the powers of sine n is an even integer). These integrals can often be simplified using power reduction identities.

When powers of a trigonometric ratios are even, the double angle identities becomes very useful. These identities reduce the even powers of sine and cosine to sums of trigonometric functions of lower powers. This process makes the integration easier. These identities includes:

$$cos^2 \theta = \frac{1}{2}(1+cos2\theta)$$
$$sin^2 \theta = \frac{1}{2}(1-cos2\theta)$$

Example problem

$$\int cos^4\theta d \theta = \int cos^2 \theta cos^2\theta = \int \frac{1}{2}(1+cos2\theta)\frac{1}{2}(1+cos2\theta)d\theta$$
$$=\int \frac{1}{4}(1)(1+2cos2 \theta)+cos2 \theta(1+cos2 \theta)d\theta$$
$$=\int \frac{1}{4}(1+cos2\theta+cos2\theta + cos^22\theta)d\theta $$
$$=\int \frac{1}{4}(1+2cos2\theta+ cos^22\theta)d\theta $$
$$=\frac{1}{4} \int 1 + 2cos2\theta + \frac{1}{2}(1 + cos4\theta)d\theta$$
$$=\frac{1}{4}\int d\theta + \frac{1}{2} \int cos2\theta + \frac{1}{8} \int d\theta + \frac{1}{8} \int cos 4 \theta d\theta $$
$$=\frac{1}{4}\theta + \frac{1}{2} \frac{sin2\theta}{2} +\frac{1}{8}\theta +\frac{1}{8} \frac{sin4\theta}{4} +C $$
$$=\frac{\theta}{4} + \frac{sin2\theta}{4} + \frac{\theta}{8}+\frac{sin4\theta}{32} + C $$

simplifying:

$$=\frac{3\theta}{8} +\frac{sin2\theta}{4} +\frac{sin4\theta}{32} + C $$

Example Problem 2

solve:

$$ \int sin^24\theta d\theta $$

solution

using double angle identities;

$$sin^24\theta = \frac{1}{2}(1-cos8\theta)d\theta $$

and so:

$$\int sin^24\theta d\theta = \int \frac{1}{2}(1-cos8\theta)d\theta = \frac{1}{2}\int d\theta -\frac{1}{2}\int cos 8\theta d\theta $$

integrating we find :

$$=\frac{1}{2}\theta – \frac{1}{16}sin8\theta + C =\frac{\theta}{2} -\frac{sin8\theta}{16} + C$$

Solving Integrating trigonometric expressions with odd powers

while integrating this kind of trigonometric expressions, we use trigonometric identity which states that:

$$cos^2\theta + sin^2\theta = 1$$
Right angled triangle to explain Integrating trigonometric expressions

consider the integral:

$$\int sin^5\theta d\theta$$

we split the operation sin5θ into sin4θ(sinθ).

thus:

but sin4θ = (1-cos2θ)2 from the identity stated earlier.

Expanding:

$$(1-cos^2θ)^2 = (1-2cos^2\theta +cos^4\theta)θdθ$$

it follows that:

$$\int sin^4θ sinθ dθ = \int (1-cos^2θ)^2sinθdθ$$
$$= \int (1-2cos^2θ -cos^4θ)sinθdθ$$
$$=\int (sinθ – 2sinθcos^2θdθ + sinθcos^4θ)dθ$$
$$=\int sinθdθ – 2 \int sinθcos^2θdθ + \int sinθcos^4θdθ$$

remember the identity:

$$ \int sinθcos^nθdθ = \frac{-1}{n+1}cos^{n+1}θ+c$$

hence

$$\int sinθcos^4θdθ = \frac{-1}{5}cos^5θ +C$$

similarly:

$$\int sinθcos^2θdθ = \frac{-1}{3}cos^3θ +C$$

hence,

$$=\int sinθdθ – 2 \int sinθcos^2θdθ + \int sinθcos^4θdθ = -cos\theta + \frac{2}{3}cos^3\theta – \frac{1}{5}cos^5\theta + C$$

In general, integrating even power of sine and cosine constitutes the following procedure:

  • If the integrand contains an even power of sine or cosine, use the corresponding power reduction identity. This will simplify the expression.
  • After applying the identity, the integral often becomes easier to solve. This is because the powers of sine or cosine are reduced to terms involving cos⁡(2x) or constants. These terms can be integrated straightforwardly.
  • Repeat where necessary: If higher powers of sine or cosine appear, you may need to apply the power reduction identity. You might have to do this multiple times.

Example 3

solve:

$$\int cos^3 \theta d\theta $$

solution

$$cos^3θ = cos^2θcosθ = (1-sin^2θ)cosθdθ = (cosθ-sin^2θcosθ)dθ$$

hence:

$$\int cos^3 \theta d\theta =\int(1-sin^2θ)cosθdθ =\int (cosθ-sin^2θcosθ)dθ$$
$$\int (cosθ-sin^2θcosθ)dθ = \int cosθdθ-\int sin^2θcosθdθ$$
$$ \int cosθdθ-\int sin^2θcosθdθ = sinθ -\frac{1}{3}sin^3θ+C$$

Example 3

solve:

$$\int tan^3 \theta d\theta $$

solution

incase of tanθ, we use the identity :

$$1 + tan^2θ = sec^2θ$$
$$\int tan^3 \theta d\theta = \int tan^2θtanθdθ = \int(sec^2θ-1)tanθdθ$$
$$ \int(sec^2θ-1)tanθdθ = \int (sec^2θtanθ-tanθ)dθ$$
$$ = \int sec^2θtanθdθ-\int tanθdθ$$

we can use integration by substitution to solve

$$ \int sec^2θtanθdθ$$

let u = tanθ

$$\frac{du}{d\theta} = sec^2\theta d\theta$$

hence du = sec2θdθ

$$dθ = \frac{du}{sec^2θ}$$
$$\int tanθ {sec^2θ}\frac{du} {{sec^2θ}} =\int tanθ du= \int u du$$

and

$$ \int u du = \frac{u^2}{2}+C = \frac{tan^2\theta}{2} +c$$

and

$$\int tanθdθ = ln|cosθ|+c$$

hence

$$ = \int sec^2θtanθdθ-\int tanθdθ =\frac{tan^2\theta}{2} +ln|cosθ|+c$$

Example 4

$$\int sec^6 2xdx$$

solution to example 4

$$\int sec^6 2xdx = \int sec^42x sec^2xdx = \int (1+tan^2x)^2sec^22xdx \ \ ——(i)$$

that is:

$$sec^22x = 1 + tan^22x$$ $$sec^42x = (sec^22x)^2 = (1+tan^22x)^2$$

Expanding the equation (i) above:

$$\int (sec^22x + 2tan^22xsec^22x + tan^42xsec^22x)dx$$

which results to:

$$\int sec^22x dx +2 \int tan^22xsec^22x dx+ \int tan^42xsec^22x dx$$

consider the integral;

$$\int sec^22x dx$$

let u=2x

then:

$$\frac{dx}{du} = \frac{1}{2}$$
$$dx = \frac{1}{2} du$$

Therefore:

$$dx = {2du}$$
$$\int sec^22 x dx = \frac{1}{2} \int sec^2 u dx = \frac{1}{2} tan u + C = \frac{tan 2x} {2} + C \ ……………(ii)$$

consider the expression:

$$2 \int tan^2 2x sec^2 2x dx$$

let u = tan2x

du = 2sec22xdx

$$dx = \frac{du} {2sec^2 2x}$$

the expression therefore becomes:

$$2 \int tan^2 2x sec^2 2x dx = 2 \int u^2 sec^2 2x \frac{du} {2sec^2 2x} $$ $$= \int u^2 du = \frac{u^3} {3} +C = \frac{tan^32x} {3} +c \ ………(iii)$$

Using the same method and procedure:

$$\int tan^42xsec^22x dx = \frac{tan^52x} {10} + C \ ……………………..(iv)$$

combining the results of (ii), (iii) and (iv), the integral becomes;

$$\int sec^6 2xdx = \frac{tan 2x} {2} + \frac{tan^32x} {3} + \frac{tan^52x} {10} C $$

Multiple angles identities

Multiple angle identities are trigonometric identities that express trigonometric functions of multiple angles (such as 2θ, 3θ, etc.) in terms of simpler functions of a single angle θ. These identities are helpful in simplifying expressions involving trigonometric functions of multiple angles.

Some common multiple angles includes:

  • sin(A+B) +sin(A-B) = 2sinAcosB
  • cos(A+B) + cos(A-B) = 2cosAcosB
  • cos(A-B)-cos(A+B) = 2sinAsinB

Example

work out:

$$\int sin5 \theta cos3 \theta d\theta$$

the solution to this would be:

this follows from: sin(A+B) +sin(A-B) = 2sinAcosB

$$sinAcosB = \frac{1}{2}[sin(A+B) + sin(A-B)]$$

in our case: A=5θ and B = 3θ

hence:

$$sin5θcos3θ= \frac{1}{2}[sin(5θ+3θ) + sin(5θ-3θ)]$$ $$ =\frac{1}{2}[sin(8θ) + sin(2θ)]=\frac{1}{2}sin 8θ +\frac{1}{2}sin 2θ $$

therefore:

$$\int sin5 \theta cos3 \theta d\theta = \frac{1}{2}\int (sin8 \theta + sin2 \theta) d \theta$$
$$\int sin 8θ = \frac{1}{8}cos8θ$$

and

$$\int sin 2θ = \frac{1}{2}cos8θ$$

it follows that:

$$ = \frac{1}{2}\int (sin8 \theta + sin2 \theta) d \theta = \frac{1}{16} cos 8 \theta+\frac{1}{4}cos2 \theta + C $$

Example 2

work out:

$$\int sin3θ cos^2θ dθ $$

you can proceed as follow:

from double angle identities:

$$cos^2θ = \frac{1+cos2θ }{2} dθ $$

therefore

$$\int sin3θ (\frac{1+cos2θ }{2} )dθ =\frac{1}{2} \int sin3θ +\frac{1}{2} \int sin3θ cos2θ$$

but

$$sin3θ cos2θ =\frac{1}{2} (sin 5θ+ sinθ)$$
$$\frac{1}{2} \int sin3θ +\frac{1}{2} \int sin3θ cos2θ = \frac{1}{2} \int sin3θdθ+\frac{1}{2} \int \frac{1}{2} (sin5θ+sinθ)dθ$$
$$= \frac{1}{2} \int sin3θdθ+\frac{1}{4} \int (sin5θ+sinθ)dθ$$
$$= \frac{1}{2} \int sin3θdθ+\frac{1}{4} \int sin5θ+ \frac{1}{4} \int sinθdθ$$
$$=- \frac{1}{6}cos3θ-\frac{1}{20}cos5θ-\frac{1}{4}cosθ +C$$

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