Differentiation in calculus can be defined as instantaneous rate of change of a variable with respect to another variable present in the function. It is also called as finding out the derivative of a function. If we consider tangent line to the function then the slope of that tangent line at the point of contact is known as the differentiation of the function.

Finding out derivative of a function involves number of methods. One such method is the substitution method. This method is used to find the derivative of complex functions which is simplified by substituting a part of that function as something which simplifies the function and makes it easy to differentiate. It transforms the function into standard forms where standard formulas of differentiation can be applied. This method of differentiation can either be applied directly or in the chain rule of differentiation. The chain rule of differentiation formula is as follows. In order to differentiate a function of a function, 

$y$ = $f[g(x)]$, that is to find $\frac{dy}{dx}$ we need to do two things:

Substitute $u$ = $g(x)$, this gives us $y$ = $f(u)$. After that we apply a formula known as chain rule

Chain Rule: $\frac{dy}{dx}$ = $\frac{dy}{du}$ $\times$ $\frac{du}{dx}$
We can substitute functions while carrying out the differentiation of complex functions. It is done to simplify the proce\beta of finding out derivative. Few of the functions which could be differentiated using substitution method are as follows:

If $f(x)$ contains terms like $\sqrt{(a^2 - x^2)}$, we can substitute the value of $x$ = $a$ sin $t$ or $x$ = $a$ cos $t$ to simplify the function to be differentiated

If $f(x)$ contains terms like $\sqrt{(a^2 + x^2)}$, we can substitute the value of $x$ = $a$ tan $t$ or $x$ = $a$ cot $t$ to simplify the function to be differentiated

If $f(x)$ contains terms like $\sqrt{(x^2 - a^2)}$, we can substitute the value of $x$ = $a$ sec $t$ or $x$ = $a$ csc $t$ to simplify the function to be differentiated

If $f(x)$ contains terms like $\sqrt{(a - x)}$ or $\sqrt{(a + x)}$, we can substitute the value of $x$ = $a$ cos $2t$ to simplify the function to be differentiated

If $f(x)$ contains terms like $\sqrt{(a^2 - x^2)}$, we can substitute the value of $x^2$ = $a^2$ cos $2t$ to simplify the function to be differentiated

If $f(x)$ contains terms like a sin $x + b$ cos $x$, we can substitute the value of $a$ = $r$ cos $\alpha$ and $b$ = $r$ sin $\alpha$ to simplify the function to be differentiated

If $f(x)$ contains terms like $\sqrt{(x - \alpha)}$ and $\sqrt{(\beta - x)}$, we can substitute the value of $x$ = $\alpha\ sin^2\ t + \beta\ cos^2\ t$ to simplify the function to be differentiated

If $f(x)$ contains terms like $\sqrt{(2ax - x^2)}$, we can substitute the value of $x$ = $a (1 - cos\ t)$ to simplify the function to be differentiated.
Integration by substitution is a method to simplify the technique of integrating complex integrals. Integration by substitution is also known as $u$ substitution method. It is possible to apply integration by substitution when the integral contains both the function and its derivative. The integral must be able to write it in the following form:

$\int\ f\ [g(x)]\ g'(x)\ dx$

Then we can substitute $g(x)$ = $u$ and $g'(x)$ = $du$, giving us

$\int\ f(u)\ du$

And then after integrating, at the last step re insert $g(x)$ wherever $u$ lies

There are certain integration by substitution formulas which could be applied to make the integration technique easier. Few of them are as follows:

If an integration is in the form $\frac{\int\ 1}{\sqrt{ (a^2 - x^2)}}$ $dx$ then by substituting the value of $x$ = $a$ sin $t$ or $x$ = $a$ cos $t$, we get $\frac{\int\ 1}{\sqrt{(a^2 - x^2)}}$ $dx$ = $sin^{-1}$ $(\frac{x}{a})$ + $c$

If an integration is in the form $\frac{\int\ 1}{\sqrt{ (a^2 + x^2)}}$ $dx$ then by substituting the value of $x$ = $a$ tan $t$ or $x$ = $a$ cot $t$, we get $\frac{\int\ 1}{\sqrt{(a^2 + x^2)}}$ $dx$ = $sinh^{-1}$ $(\frac{x}{a})$ + $c$

If an integration is in the form $\frac{\int\ 1}{\sqrt{(x^2 - a^2)}}$ $dx$ then by substituting the value of $x$ = $a$ sec $t$ or $x$ = $a$ csc $t$, we get $\frac{\int\ 1}{\sqrt{(x^2 - a^2)}}$ $dx$ = $cosh^{-1}$ $(\frac{x}{a})$ + $c$

If an integration is in the form $\frac{\int\ a}{x \times \sqrt{(x^2 - a^2)}}$ $dx$ then by substituting the value of $x$ = $a$ sec $t$ or $x$ = $a$ csc $t$, we get $\frac{\int\ a}{x \times \sqrt{(x^2 - a^2)}}$ $dx$ = $sec^{-1}$ $(\frac{x}{a})$ + $c$

If an integration is in the form $\frac{\int\ a}{x \times \sqrt{(a^2 - x^2)}}$ $dx$ then by substituting the value of $x$ = $a$ sin $t$ or $x$ = $a$ cos $t$, we get $\frac{\int\ a}{x \times \sqrt{(a^2 - x^2)}}$ $dx$ = $- sech^{-1}$ $(\frac{x}{a})$ + $c$

If an integration is in the form $\frac{\int\ a}{x \times \sqrt{(a^2 + x^2)}}$ $dx$ then by substituting the value of $x$ = $a$ tan $t$ or $x$ = $a$ cot $t$, we get $\frac{\int\ a}{x \times \sqrt{(a^2 + x^2)}}$ $dx$ = $- csch^{-1}$ $(\frac{x}{a})$ + $c$

If an integration is in the form $\frac{\int\ a}{(a^2 + x^2)}$ $dx$ then by substituting the value of $x$ = $a$ tan $t$ or $x$ = $a$ cot $t$, we get $\frac{\int\ a}{(a^2 + x^2)}$ $dx$ = $tan^{-1}$ $(\frac{x}{a})$ + $c$

If an integration is in the form $\frac{\int\ a}{(a^2 - x^2)}$ $dx$ then by substituting the value of $x$ = $a$ sin $t$ or $x$ = $a$ cos $t$, we get $\frac{\int\ a}{(a^2 - x^2)}$ $dx$ = $tanh^{-1}$ $(\frac{x}{a})$ + $c$

If an integration is in the form $\frac{\int\ a}{(x^2 - a^2)}$ $dx$ then by substituting the  value of $x$ = $a$ sec $t$ or $x$ = $a$ csc $t$, we get $\frac{\int\ a}{(x^2 - a^2)}$ $dx$ = $coth^{-1}$ $(\frac{x}{a})$ + $c$
Example 1: 

Differentiate the given function using substitution method

$Y$ = $tan^{-1}$ $\frac{\sqrt{(1 + x^2) – 1}}{x}$

Solution: 

The given function $Y$ = $tan^{-1}$ $\frac{\sqrt{(1 + x^2)} – 1}{x}$ is first simplified using the substitution, $x$ = tan $t$. Differentiating with respect to $t$, $\frac{dx}{dt}$ = $sec^2\ t$ or $\frac{dt}{dx}$ = $\frac{1}{sec^2\ t}$ = $cos^2\ t$

$Y$ = $tan^{-1}$ $\frac{\sqrt{(1 + tan^2t) – 1}}{tan\ t}$

$Y$ = $tan^{-1}$ $\frac{\sqrt{sec^2 t – 1}}{tan\ t}$

$Y$ = $tan^{-1}$ $\frac{sec t – 1}{tan\ t}$

$Y$ = $tan^{-1}$ $\frac{\frac{1\ -\ cos\ t}{cos\ t}}{\frac{sin\ t}{cos\ t}}$

$Y$ = $tan^{-1}$ $\frac{1 – cos\ t}{sin\ t}$

$Y$ = $tan^{-1}$ $2sin^2$ $\frac{(\frac{t}{2})}{2sin}$ $(\frac{t}{2})$ cos $(\frac{t}{2})$

$Y$ = $tan^{-1}$ $tan$ $(\frac{t}{2})$

$Y$ = $\frac{t}{2}$

Differentiating with respect to $t$, we get

$\frac{Dy}{dt}$ = $\frac{1}{2}$

$\frac{Dy}{dx}$ = $\frac{dy}{dt}$ $\times$ $\frac{dt}{dx}$

$\frac{Dy}{dx}$ = $\frac{1}{2}$ $\times\ cos^2\ t$

$\frac{Dy}{dx}$ = $\frac{cos^2\ t}{2}$

If $tan\ t$ = $x \Rightarrow\ tan^2\ t$ = $x^2\ \Rightarrow\ sec^2\ t$ = $1 + x^2\ \Rightarrow\ cos^2\ t$ = $\frac{1}{(1 + x^2)}$

Thus, $\frac{dy}{dx}$ = $\frac{1}{2}$ $(1 + x^2)$
Example 2:

Integrate the following functions using the substitution method:

a) $\frac{\int\ tan\ x}{sec^4\ x\ dx}$

b) $\frac{\int\ dx}{1 + e^x\ dx}$

c) $\int\ 2x\ (x^2 + 1)^3\ dx$

Solution: 

a) $\frac{\int\ tan\ x}{sec^4\ x\ dx}$

It is possible to apply integration by substitution when the integral contains both the function and its derivative. So, to get that form we multiply both the numerator and denominator by sec $x$.

$\frac{\int\ sec\ x\ tan\ x}{sec^5\ x\ dx}$

Let us substitute sec $x$ = $u$, then on differentiating we get sec $x$ tan $x\ dx$ = $du$. The integral now looks like,

$\frac{\int\ du}{u^5}$

= $\int\ u^{-5}\ du$

= $u^{\frac{(-5 + 1)}{(-5 + 1)}}$ + $c$

= $u^{\frac{-4}{-4}}$ + $c$

= $\frac{- 1}{4u^4}$ + $c$

Plugging in back the value of $u$ in the solution we get,

= $\frac{- 1}{sec^4}$ $x + c$

b) $\frac{\int\ dx}{1}$ + $e^x\ dx$

It is possible to apply integration by substitution when the integral contains both the function and its derivative. So, to get that form we multiply both the numerator and denominator by $e^x$

$\frac{\int\ e^x\ dx}{e^x(1 + e^x)\ dx}$

Let us substitute $e^x$ = $u$, then on differentiating we get $e^x\ dx$ = $du$. The integral now looks like,

= $\frac{\int\ du}{u(1 + u)}$

= $\frac{\int\ (1 + u)\ –\ u\ du}{u(1 + u)}$

= $\frac{\int\ du}{u}$ - $\frac{\int\ du}{(1 + u)}$

= $ln |u| - ln |1 + u| + c$

= $ln$ $|\frac{u}{1}$ + $u|\ +\ c$

Plugging in back the value of $u$ in the solution we got,

= $ln$ $|\frac{e^x}{1}$ + $e^x|\ +\ c$

c) $\int\ 2x\ (x^2 + 1)^3\  dx$

Let us substitute $x^2 + 1$ = $u$, then on differentiating we get $2x\ dx$ = $du$

= $\int\ u^3\ du$

= $u^{\frac{(3 + 1)}{(3 + 1)}}$ + $c$

= $\frac{u^4}{4}$ + $c$

= $\frac{(x^2 + 1)}{4}$ + $c$