Trigonometry and calculus are two most useful branches of mathematics. The applications of those are seen almost everywhere. Though, differentiation and integration are the concepts of calculus, but these, especially integration, can never be understood clearly without a good command over trigonometry. There are various complex looking integrands that cannot be evaluated using usual formulas. These are integrated using trigonometric substitutions.
Trigonometric substitution is the process of plugging trigonometric functions into other expressions which are needed to be integrated. Trigonometric substitutions actually make the expression easier to integrate.
In order to understand this method more clearly, let us learn trig substitution method in detail.

There are few integrands that can not be solved by usual methods. These integrands are such a way that these are solved by plugging particular trigonometric substitutions.
In order to perform an integration using trigonometric substitution, we need to follow the steps mentioned below:
Step 1: First we need to recognize the proper integrand which can be solved using trigonometric substitution.
Step 2: Determine appropriate trigonometric expression that has to be substituted in place of some other expression in integrand.
Step 3: Differentiate the substitution expression and find the value of dx (where, x is independent variable with respect to which the integration is performed).
Step 4: Substitute the values in given integrand.
Step 5: Use a suitable trigonometric identity to simplify the expression.
Step 6: Now, we will be able to integrate the expression with respect to new substituted variable using the familiar integration formulas.
Step 7: Substituting back in the value of assumed variable.
Let us have a look at a simple example in order to understand the process of trigonometric substitution.
Example: Solve $\int$$\frac{1}{\sqrt{a^{2}-x^{2}}}$$dx$

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

Substitute $x$ = $a sin\theta$

Then $dx$ = $a cos\theta\ d\theta$ .......(Differentiating our assumption)

= $\int$$ \frac{1}{\sqrt{a^{2}-(a sin\theta)^{2}}}$ $a cos \theta\ d\theta$

= $\int$$ \frac{1}{\sqrt{a^{2}-a^{2} sin^{2} \theta}}$ $a cos \theta\ d\theta$
= $\int $$\frac{1}{\sqrt{a^{2}(1- sin^{2} \theta)}}$ $a cos \theta\ d\theta$
= $\int$ $\frac{1}{a \sqrt{1- sin^{2} \theta\ }}$ $a cos \theta\ d\theta$

= $\int$$ \frac{1}{a \sqrt{cos^{2} \theta}}$ $a cos \theta\ d\theta$ ......$(1-sin^{2}\theta=cos^{2}\theta)$

= $\int$ $\frac{1}{a cos \theta}$ $a cos \theta\ d\theta$

= $\int d\theta$

= $\theta$ + $C$ .........( C = integration constant)

substituting back the value of $\theta$

= $sin^{-1}$$(\frac{x}{a})$ +$C$
There are three main types of trigonometric substitutions that are used in order to evaluate the integration. These are listed below in the table :
Identity used
   $a^{2}-x^{2}$      $x$ = $a sin\theta$  
$x$ = $a tan\theta$
$x^{2}-a^{2}$  $x$ = $a sec\theta$
In the first column, the integrand is written. If same format of expression occurs, then we must substitute the trigonometric substitution that is mentioned in the corresponding second column in front of each integrand. Third column is for identity that is used after substitution is accomplished.
In the process of integration by trigonometric substitution, we replace the independent variable by another variable which is our assumption.
After substituting trig function, we need to integrate the expression with respect to another variable which was not at all present in the given question.
Therefore, in order to give answer in terms of given variable, we have to substitute back the value of assumed variable in terms of given variable. This value in an inverse trigonometric function.
For example: Let us consider that in order to solve an arbitrary problem, one has assumed $x$ = $a tan \theta$
Where $x$ is the independent variable given in the question, while $\theta$ is assumed.
Hence at the end, one needs to put the value of $\theta$ back in the solved expression in order to get to the right answer.
From above assumption:
$x$ = $a tan\theta$

$\frac{x}{a}$=$tan \theta$

Few examples based on integration using trigonometric substitution are illustrated below:
Example 1: Integrate $\frac{1}{x^{2}\sqrt{x^{2}-9}}$ with respect to $x$.

Solution: $\int$ $\frac{1}{x^{2} \sqrt{x^{2}-9}}$$dx$

= $\int$ $\frac{1}{x^{2} \sqrt{x^{2}-3^{2}}}$$dx$

Let us assume that $x$ = $3 sec t$

Then, $dx$ = $3 tan t \sec t dt$

Substituting above values, we get

= $\int$ $\frac{1}{(3sec\ t)^{2} \sqrt{(3sec\ t)^{2}-3^{2}}}$$dx$

= $\int$ $\frac{1}{9sec^{2}t \sqrt{9sec^{2}t-9}}$$3 tan\ t\ sec\ t\ dt$

= $\int $ $\frac{1}{27sec^{2}t  \sqrt{sec^{2}t-1}}$$3 tan\ t\ sec\ t\ dt$

= $\frac{1}{9} \int \frac{1}{sec^{2}t  tan\ t}$$ tan\ t\ sec\ t\ dt$

= $\frac{1}{9} $$\int cos\ t\ dt$

= $\frac{1}{9}$ $sin\ t + C$

= $\frac{1}{9}$ $sin$  $($$sec^{-1}$$\frac{x}{3}$$)$ + $C$ (answer)
the value of sin t can also be evaluated from assumed relation
$x$ = $3 sec t$

$sec t$ = $\frac{x}{3}$

so, $sin t$ = $\frac{\sqrt{x^{2}-9}}{x}$
$\frac{1}{9}$ $sin\ t + C$ = $\frac{1}{9} \frac{\sqrt{x^{2}-9}}{x}$ + $C$ (answer)

Example 2: Evaluate $\int $$\frac{1}{\sqrt{x^{2}+4}}$$dx$

Solution: $\int$ $\frac{1}{\sqrt{x^{2}+4}}$$dx$

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

Let us substitute $x$ = $2 tan \theta$

$dx$ = $2 sec^{2}\theta\ d\theta$

= $\int$$ \frac{1}{\sqrt{(2 tan \theta)^{2}+2^{2}}}$$2 sec^{2}\theta\ d\theta$

= $\int $$\frac{1}{\sqrt{4 tan^{2}\theta+4}}$$2 sec^{2}\theta\ d\theta$

= $\int $$\frac{1}{\sqrt{4 (tan^{2}\theta+1)}}$$2 sec^{2}\theta\ d\theta$

= $\int$ $\frac{1}{\sqrt{ sec^{2}\theta}}$$ sec^{2}\theta\ d\theta$

= $\int $$\frac{1}{ sec\theta} $$sec^{2}\theta\ d\theta$

= $\int sec\theta\ d\theta$

= $tan\theta\ sec\theta+C$

Since, $x$= $2 tan \theta$ or $tan\theta$=$\frac{x}{2}$

Hence, $sec\theta$=$\frac{\sqrt{x^{2}+4}}{2}$

Substituting back the values of $tan\theta$ and $sec\theta$ in the evaluated result.

$\int$$ \frac{1}{\sqrt{x^{2}+4}}$$dx$ = $\frac{x}{2}\frac{\sqrt{x^{2}+4}}{2}$+$C$

= $\frac{x\sqrt{x^{2}+4}}{4}$+$C$ (answer).