Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# Integration Using Substitution

Integration is a method explained under calculus, apart from differentiation, where we find the integrals of functions. Integration by substitution is one of the methods to solve integrals. This method is also called u-substitution. Also, find integrals of some particular functions here.

The integration of a function f(x) is given by F(x) and it is represented by:

∫f(x)dx = F(x) + C

Here R.H.S. of the equation means integral of f(x) with respect to x.

F(x) is called anti-derivative or primitive.

f(x) is called the integrand.

dx is called the integrating agent.

C is called constant of integration or arbitrary constant.

x is the variable of integration.

The anti-derivatives of basic functions are known to us. The integrals of these functions can be obtained readily. But this integration technique is limited to basic functions and in order to determine the integrals of various functions, different methods of integration are used. Among these methods of integration let us discuss integration by substitution.

## Integration By Substitution Method

In this method of integration, any given integral is transformed into a simple form of integral by substituting the independent variable by others.

Take for example an equation having independent variable in x , i.e.

$$\begin{array}{l}\int \sin (x^{3}).3x^{2}.dx\end{array}$$
———————–(i),

In the equation given above the independent variable can be transformed into another variable say t.

Substituting

$$\begin{array}{l} x^{3} = t \end{array}$$
———————-(ii)

Differentiation of above equation will give-

$$\begin{array}{l}3x^{2}.dx = dt\end{array}$$
———————-(iii)

Substituting the value of (ii) and (iii) in (i), we have

$$\begin{array}{l}\int \sin (x^{3}).3x^{2}.dx = \int \sin t . dt\end{array}$$

Thus the integration of the above equation will give

$$\begin{array}{l}\int \sin t .dt = -\cos t + c \end{array}$$

Again putting back the value of t from equation (ii), we get

$$\begin{array}{l}\int \sin (x^{3}).3x^{2}.dx = -\cos (x^{3}) + c\end{array}$$

The General Form of integration by substitution is:

$$\begin{array}{l}\int f(g(x)).g'(x).dx = f(t).dt\end{array}$$
,

where t = g(x)

Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function. To understand this concept better let us look into an example.

 Example: Find the integration of $$\begin{array}{l}\int \frac{e^{\tan^{-1}x}}{1+x^{2}}.dx\end{array}$$. Solution: Given  $$\begin{array}{l}\int \frac{e^{\tan^{-1}x}}{1+x^{2}}.dx\end{array}$$ Let $$\begin{array}{l}t = \tan^{-1}x\end{array}$$…………….(i) $$\begin{array}{l}\Rightarrow dt = \frac{1}{1+x^{2}}.dx\end{array}$$ $$\begin{array}{l}I = \int e^{t}. dt\end{array}$$ $$\begin{array}{l}= e^{t} + C\end{array}$$     ……………….(ii) Substituting the value of (i) in (ii), we have $$\begin{array}{l}I = e^{\tan^{-1}x} + C\end{array}$$ Example: Integrate $$\begin{array}{l}2x \cos(x^{2}-5)\end{array}$$ with respect to x. Solution: $$\begin{array}{l}I = \int 2x \cos(x^{2}-5) .dx\end{array}$$ Let $$\begin{array}{l}x^{2} – 5 = t\end{array}$$    …………….(i) $$\begin{array}{l}\Rightarrow 2x .dx= dt\end{array}$$ Substituting these values, we have $$\begin{array}{l}I = \int \cos(t).dt\end{array}$$ $$\begin{array}{l}= \sin t + C\end{array}$$   ……………..(ii) Substituting the value of (i) in (ii), we have $$\begin{array}{l}= \sin(x^{2}- 5) + C\end{array}$$ This is the required integration for the given function.