Integration of trigonometric functions

Integrating trigonometric functions often involves using various integration techniques, including basic integration formulas, substitution, integration by parts, and trigonometric identities. Here’s a brief overview of some common integrals of trigonometric functions:

Basic Trigonometric Integrals

• $\int (\cos x) = \sin x + C$
• $\int (\sin x) = – \cos x + C$
• $\int ( \sec^2 x) = \tan x + C$
• $\int (\csc^2 x) = -\cot x + C$
• $\int ( \sec (x) \tan (x) )=\sec x + C$
• $\int (\csc (x) \cot (x)) = -\csc x + C$

Integrals of Tangent and Cotangent

This can be solved using Integration by substitution

(I)

\[
\int \cot(x) \, dx = \int \frac {\cos(x)}{\sin(x)} \, dx
\]

Now
$\sin(x) =t$
then
$\cos (x) dx = dt$
Therefore,

\[
\int \cot(x) \, dx = \int \frac {1}{t} \, dt = \ln |sin(x)| + C
\]

(II)

\[
\int \tan(x) \, dx = \int \frac {\sin(x)}{\cos(x)} \, dx
\]

Now
$\cos(x) =t$
then
$-\sin (x) dx = dt$
Therefore,

\[
\int \cot(x) \, dx = -\int \frac {1}{t} \, dt = -\ln |cos(x)| + C=ln|\sec (x)| + C
\]

Integrals of Secant and Cosecant

(I)

Integral of cosecant can be little tricky and can be solved by multiplying and dividing by $(\csc(x) – \cot(x))$. Here’s how it’s done:

\[
\int \csc(x) \, dx = \int \csc(x) \cdot \frac{\csc(x) – \cot(x)}{\csc(x) -\cot(x)} \, dx \\
= \int \frac{\csc^2(x) – \csc(x)\cot(x)}{\csc(x) – \cot(x)} \, dx.
\]

Now, taking u as
$ u= \csc(x) – \cot(x) $
$ du =(-\csc(x)\cot(x) + \csc^2(x)) dx $

Therefore integration becomes

\[
\int \csc(x) \, dx =\int \frac {1}{u} \, du,
\]

hence

\[
\int \csc(x) \, dx = \ln | \csc(x) – \cot(x) | + C
\]

(II)

Integral of secant can be little tricky and can be solved by multiplying and dividing by $(\sec(x) + \tan(x))$. Here’s how it’s done:

\[
\int \sec(x) \, dx = \int \sec(x) \cdot \frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)} \, dx \\
= \int \frac{\sec^2(x) + \sec(x)\tan(x)}{\sec(x) + \tan(x)} \, dx.
\]

Now, taking u as
$ u= \sec(x) + \tan(x) $
$ du =(\sec(x)\tan(x) + \sec^2(x)) dx $

Therefore integration becomes

\[
\int \sec(x) \, dx =\int \frac {1}{u} \, du,
\]

hence

\[
\int \sec(x) \, dx = \ln | \sec(x) + \tan(x) | + C
\]

Inverse Trigonometric Functions

Integrals that result in inverse trigonometric functions, like $\int \frac{dx}{\sqrt{1 – x^2}} = \sin^{-1}(x) + C$ or $\int \frac{dx}{1 + x^2} = \tan^{-1}(x) + C$.

$\int ( \frac {1}{\sqrt {1-x^2} } ) = \sin^{-1}x + C$

$\int (\frac {1}{\sqrt {1-x^2}}) = – \cos ^{-1}x  +C$

$\int ( \frac {1}{1 + x^2}) =\tan ^{-1}x + C$

$\int ( \frac {1}{1 + x^2}) = -\cot ^{-1}x + C$

$\int (\frac {1}{|x|\sqrt {x^-1}}) = -sec^{-1} x + C $

$\int (\frac {1}{|x|\sqrt {x^-1}}) = -cosec^{-1} x + C $

Integrals Involving Substitution:

For integrals like $\int \sin(ax+b) \, dx$ or $\int \cos(ax + b) \, dx$, a substitution such as (u = ax+b) can be used.

$\int \cos (ax+b) = \frac {1}{a}  \sin (ax+b) + C$

$\int \sin (ax+b) = – \frac {1}{a} \cos (ax+b) + C$

$\int \sec^2 (ax+b) = \frac {1}{a}  \tan (ax +b) + C$

$\int \csc^2 (ax+b) = – \frac {1}{a}  \cot (ax+b)+ C$

$ \int \tan (ax+b) =- \frac {1}{a}  ln |\cos (ax+b)| + C$

$ \int \cot (ax+b) = \frac {1}{a}  ln |\sin (ax+b)| + C$

$ \int \sec (ax+b) =\frac {1}{a} ln |\sec (ax+b) + \tan (ax+b)| + C$

$ \int \csc (ax+b) = \frac {1}{a} ln |\csc (ax+b) – \cot (ax+b)| + C$

Integrals Involving Trigonometric Identities

Sometimes, using trigonometric identities to simplify the integrand is helpful. For example, $\sin^2(x)$ can be expressed as$\frac{1 – \cos(2x)}{2}$ using the double-angle identity

(I)

$\int \sin^2 x \; dx$ or $\int \cos^2 x \; dx$
We know that
$\cos2x = \cos^{2}x-\sin^{2}x = 2\cos^{2} x-1=1-2\sin^{2}x$
So
$\cos^2 x= \frac {1+ \cos 2x}{2}$
$\sin^2 x= \frac {1- \cos 2x}{2}$

(II)

$\int \sin^3 x \; dx$ or $\int \cos^3 x \; dx$
We know that
$\sin(3x)=3\sin(x)-4\sin^{3}x$
$\sin^{3}x= \frac { 3 \sin x – \sin 3x}{4}$
Also
$\cos(3x)=4\cos^{3}x-3\cos(x)$
$\cos^{3}x= \frac {\cos 3x + 3 \cos x}{4}$

(III)

$\int \tan^2 x \; dx$ or $\int \cot^2 x \; dx$
We know that
$\sec^2x =1 + \tan^2x$
or $\tan^2x =\sec^2 x -1$
$\int \tan^2 x dx = \int (\sec^2 x -1) dx$
Now $\int ( \sec^2 x) \; dx = \tan x + C$
Therefore
$\int \tan^2 x \; dx = \int (\sec^2 x -1) dx= \tan x -x + C$
We know that
$cosec^2 x =1 + \cot^2 x$
or $ \cot^2 x = cosec^2 x -1$
$\int \cot^2 x \; dx = \int (cosec^2 x -1) \; dx$
Now $\int ( cosec^2 x) \; dx = -\cot x + C$
Therefore
$\int \cot^2 x \; dx = \int (cosec^2 x -1) \; dx= -\cot x -x + C$

Integration by Parts

This technique is useful for integrals like $\int x \sin(x) \, dx$ or $\int x \cos(x) \, dx$

$\int x \sin(x) \; dx = \sin x -x \cos(x) + C$

$\int x \cos(x) \; dx = \cos x +x \sin(x) + C$

When integrating trigonometric functions, it’s often helpful to be familiar with various trigonometric identities and properties, as they can simplify the integration process significantly. Additionally, the choice of technique depends on the specific form of the integrand.