# 3.2 METHODS OF INTEGRATION – INTEGRATION BY SUBSTITUTION

So far we have dealt with functions, either directly integrable using integration formula (or) integrable after decomposing the given functions into sums & differences.

$But\ there\ are\ functions\ like\ \frac{sin(log x)}{x},\ \frac{2x + 3}{x^2 + 3x + 5}$

which cannot be decomposed into sums (or) differences of simple functions. In these cases, using proper substitution, we shall reduce the given form into standard form, which can be integrated using basic integration formula.

When the integrand (the function to be integrated) is either in multiplication or in division form and if the derivative of one full or meaningful part of the function is equal to the other function then the integration can be evaluated using substitution method as given in the following examples.

$1.\ \int \frac{sin(log\ x)}{x}\ dx= \int sin (log\ x) \frac{1}{x}\ dx$
$Here\ \frac{d}{dx} (log\ x) = \frac{1}{x}$

The above integration can be evaluated by taking  u = log x.

$2.\ \int \frac{2x + 3}{x^2 + 3x + 5}\ dx$
$since\ \frac{d}{dx} ( x^2 + 3x + 5)\ is\ 2x + 3$

it can be integrated by  taking  u  =x2 + 3x + 5.

Integrals of the form

$\int[f(x)]^n\ f^!(x)\ dx,\ n\ \neq\ -1,$
$\int\ \frac{f^!(x)}{f(x)}\ dx$

and

$\int[F(f(x))]\ f^!(x)\ dx$

are all, more or less of the same type and the use of substitution u = f(x) will reduce the given function to simple standard form which can be integrated using integration formulae.

$\color {purple} {Example\ 1\ .}\ \color {red}{Evaluate\ :}\ \int\ (5x + 2)^6\ dx\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$put\ u = 5x + 2$
$\frac{du}{dx}= 5$
$\frac{1}{5}\ du= dx$
$\int\ (5x + 2)^6\ dx= \frac{1}{5} \int\ u^6\ du$
$=\frac{1}{5} \frac{u^7}{7} + c$
$=\frac{(5x + 2)^7}{35} + c$
$\boxed{\int\ (5x + 2)^6\ dx\ =\ \frac{(5x + 2)^7}{35} + c}$

$\color {purple} {Example\ 2\ .}\ \color {red}{Evaluate\ :}\ \int\ (x^2 + x + 1)^5\ (2x + 1) dx\ \hspace{10cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$put\ u =x^2 + x + 1$
$\frac{du}{dx}= \ 2x + 1$
$\int\ (x^2 + x + 1)^5\ (2x + 1) dx= \int\ u^5\ du$
$=\frac{u^6}{6} + c$
$=\frac{(x^2 + x + 1)^6}{6} + c$
$\boxed{\int\ (x^2 + x + 1)^5\ (2x + 1) dx\ =\ \frac{(x^2 + x + 1)^6}{6} + c}$

$\color {purple} {Example\ 3\ .}\ \color {red}{Evaluate\ :}\ \int\ (2x^2\ -\ 8 x\ +\ 5)^{11}\ (x\ -\ 2) dx\ \hspace{10cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$put\ u =\ 2x^2\ -\ 8 x\ +\ 5$
$\frac{du}{dx}= \ 4x\ -\ 8$
$\frac{du}{dx}= \ 4(x\ -\ 2)$
$\frac{1}{4}\ du\ = \ (x\ -\ 2)\ dx$
$\int\ (2x^2\ -\ 8 x\ +\ 5)^{11}\ (x\ -\ 2) dx\ =\ \frac{1}{4}\ \int\ u^{11}\ du$
$=\ {\frac{1}{4}}\ \frac{u^{12}}{12} + c$
$=\ \frac{u^{12}}{48} + c$
$=\frac{(2x^2\ -\ 8 x\ +\ 5)^{12}}{48} + c$
$\boxed{\int\ (2x^2\ -\ 8 x\ +\ 5)^{11}\ (x\ -\ 2) dx\ =\ \frac{(2x^2\ -\ 8 x\ +\ 5)^{12}}{48} + c}$

$\color {purple} {Example\ 4\ .}\ \color {red}{Evaluate:\ }\ \int\ \frac{cos\ \sqrt{x}} {\sqrt{x}}\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$put\ u\ =\ \sqrt{x}$
$\frac{du}{dx}= \ \frac{1}{2\sqrt{x}}$
$2\ du\ =\ \frac{1}{\sqrt{x}}\ dx$
$\int\ \frac{cos\ \sqrt{x}}{\sqrt{x}}\ dx\ =\ \int\ 2\ cos\ u\ du$
$=\ 2\ \ sin\ u\ + c$
$=\ 2\ sin\ \sqrt{x}\ + c$
$\color {purple} {Example\ 5\ .}\ \color {red}{Evaluate\ :}\ \int\ x^3\ (\sqrt{3\ +\ 5\ x^4})\ dx\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$put\ u =\ 3\ +\ 5\ x^4$
$\frac{du}{dx}\ = \ 20\ x^3$
$\frac{1}{20}\ du\ = \ x^3\ dx$
$\int\ x^3\ (\sqrt{3\ +\ 5\ x^4})\ dx\ =\ \int\ \frac{1}{20}\ \ u^{\frac{1}{2}}\ du$
$=\ \frac{1}{20}\ \ \int\ u^{\frac{1}{2}}\ du$
$=\ \frac{1}{20}\ \frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c$
$=\ \frac{1}{20}\ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + c$
$=\ \frac{1}{30}\ (3\ +\ 5\ x^4)^{\frac{1}{2}}\ +\ c$
$\color {purple} {Example\ 6\ .}\ \color {red}{What\ is\ the\ value\ of}\ \int\ \frac{f^! (x)}{f(x)} dx\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$put\ u\ =\ f(x)$
$\frac{du}{dx}= \ f^! (x)$
$du\ = \ f^! (x)\ dx$
$\int\ \frac{f^! (x)}{f(x)}\ dx= \int\ \frac{du}{u}$
$=\ log\ u\ +\ c$
$\boxed{\int\ \frac{f^! (x)}{f(x)}\ dx= log(f(x))\ + c}$

$\color {purple} {Example\ 7\ .}\ \color {red}{Evaluate\ :}\ \int\ cot\ x\ dx\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int\ cot\ x\ dx\ =\ \int\ \frac{cos\ x}{sin\ x}\ dx$
$put\ u\ =\ sin\ x$
$\frac{du}{dx}= \ cos\ x$
$du\ =\ cos\ x\ dx$
$\int\ \frac{cos\ x}{sin\ x}\ dx\ = \int\ \frac{du}{u}$
$=\ log\ u\ +\ c$
$= log(sin\ x)\ +\ c$
$\boxed{\int\ cot\ x\ dx\ =\ log(sin\ x) + c}$

$\color {purple} {Example\ 8\ .}\ \color {red}{Evaluate\ :}\ \int\ \frac{2x}{1 + x^2}\ dx\ \hspace{15cm}$
$put\ u\ =\ 1 + x^2$
$\frac{du}{dx}= \ 2x$

du  = 2 dx

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

=  log u + c

$= log(1 + x^2) + c$
$\boxed{\int\ \frac{2x}{1 + x^2}\ dx= log(1 + x^2) + c}$

$\color {purple} {Example\ 9\ .}\ \color {red}{Evaluate\ :}\ \int\ \frac{3x^2}{x^3\ +\ 1}\ dx\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u\ =\ x^3\ +\ 1$
$\frac{du}{dx}= \ 3x^2$
$du\ =\ 3x^2\ dx$
$\int\ \frac{3x^2}{x^3\ +\ 1}\ dx= \int\ \frac{du}{u}$
$=\ log\ u\ +\ c$
$= log(x^3\ +\ 1) + c$
$\boxed{\int\ \frac{3x^2}{x^3\ +\ 1}\ dx= log(x^3\ +\ 1) + c}$

$\color {purple} {Example\ 10\ .}\ \color {red}{Evaluate\ :}\ \int\ \frac{(2x + 1)}{ x^2 + x + 1}\ dx\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u\ =\ x^2 + x + 1$
$\frac{du}{dx}= \ 2x + 1$
$du = (2x + 1) dx$
$\int\ \frac{(2x+ 1)}{x^2 + x + 1}\ dx= \int\ \frac{du}{u}$
$=log\ u + c$
$=log( x^2 + x + 1) + c$
$\boxed{\int\ \frac{(2x+ 1)}{x^2 + x + 1}\ dx\ =\ log ( x^2 + x + 1) + c}$

$\color {purple} {Example\ 11\ .}\ \color {red}{Evaluate\ :}\ \int\ \frac{2x + 9}{ x^2 + 9x + 30}\ dx\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u\ =\ x^2 + 9x + 30$
$\frac{du}{dx}= \ 2x + 9$
$du = (2x + 9) dx$
$\int\ \frac{2x+ 9}{x^2 + 9x + 30}\ dx= \int\ \frac{du}{u}$
$=log\ u + c$
$=log( x^2 + 9x + 30) + c$
$\boxed{\int\ \frac{2x + 9}{ x^2 + 9x + 30}\ dx\ =\ log ( x^2 + 9x + 30) + c}$

$\color {purple} {Example\ 12\ .}\ \color {red}{Evaluate\ :}\ \int\ \frac{e^x}{e^x\ +\ 5}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u\ =\ e^x\ +\ 5$
$\frac{du}{dx}= \ e^x$
$du\ =\ e^x\ dx$
$\int\ \frac{e^x}{e^x\ +\ 5}\ dx= \int\ \frac{du}{u}$
$=\ log\ u\ + c$
$=\ log(e^x\ +\ 5)\ + c$
$\boxed{\int\ \frac{e^x}{e^x\ +\ 5}\ dx\ =\ log(e^x\ +\ 5)\ + c}$

$\color {purple} {Example\ 13\ .}\ \color {red}{Evaluate\ :}\ \int\ \frac{sin\ x}{1\ -\ cos\ x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u\ =\ 1\ -\ cos\ x$
$\frac{du}{dx}= \ sin\ x$
$du\ =\ sin\ x\ dx$
$\int\ \frac{sin\ x}{1\ -\ cos\ x}\ dx\ = \int\ \frac{du}{u}$
$=\ log(1\ -\ cos\ x)\ + c$
$\boxed{\int\ \frac{sin\ x}{1\ -\ cos\ x}\ =\ log(1\ -\ cos\ x)\ + c}$
$\color {purple} {Example\ 14\ .}\ \color {red}{Evaluate\ :}\ \ \int\ \frac{dx}{x\ log x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int\ \frac{dx}{x\ log x} = \int\ \frac{1}{log x}\ \frac{1}{x}\ dx$
$Put\ u\ =\ log x$
$\frac{du}{dx}=\frac{1}{x}$
$du = \frac{1}{x}\ dx$
$\int\ \frac{dx}{x\ log x}\ dx= \int\ \frac{du}{u}$
$=\ log u\ +\ c$
$=\ log ( log x )\ +\ c$
$\boxed{\int\ \frac{dx}{x\ log x}\ dx= log ( log x) + c}$

$\color {purple} {Example\ 14\ .}\ \color {red}{Evaluate\ :}\ \ \int\ \frac{sec^2(log x)}{ x}\ dx\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$

Put    u  = log x

$\frac{du}{dx} = \frac{1}{x}$
$du = \frac{1}{x}\ dx$
$\int\ \frac{sec^2(log x)}{ x}\ dx = \int sec^2\ u\ du$

=  tan u  + c

=   tan ( log x ) + c

$\boxed{\int\ \frac{sec^2(log x)}{ x}\ dx\ =\ tan ( log\ x) + c}$

$\color {purple} {Example\ 15\ .}\ \color {red}{Evaluate\ :}\ \ \int\ sin^3x\ cos x dx\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$

Put    u  =  sin x

$\frac{du}{dx} = cos x$

du  =  cos x  dx

$\int\ sin^3x\ cos x dx = \int u^3\ du$
$=\frac{u^4}{4} + c$
$=\frac{sin^4 x}{4} + c$
$\boxed{\int\ sin^3x\ cos x\ dx\ =\ \frac{sin^4 x}{4} + c}$

$\color {purple} {Example\ 16\ .}\ \color {red} {Evaluate\ :}\ \int\ e^{sin^2\ x}\ sin\ 2x\ dx\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int\ e^{sin^2\ x}\ sin\ 2x\ dx\ =\ \int\ e^{sin^2\ x}\ 2\ sin\ x\ cos\ x\ dx$
$put\ u\ =\ sin^2\ x$
$\frac{du}{dx}\ =\ 2\ sin\ x\ cos\ x$
$du\ =\ 2\ sin\ x\ cos\ x\ dx$
$\int\ e^{sin^2\ x}\ sin\ 2x\ dx = \int e^u\ du$
$=\ e^u\ + c$
$=\ e^ {sin^2\ x} + c$
$\boxed{\int\ e^{sin^2\ x}\ sin\ 2x\ dx\ =\ e^ {sin^2\ x} + c}$

$\color {purple} {Example\ 17\ .}\ \color {red} {Evaluate\ :}\ \int\ sec^7\ x\ tan\ x\ dx\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int\ sec^7\ x\ tan\ x\ dx = \int sec^6\ x\ sec\ x\ tan\ x\ dx$
$put\ u\ =\ sec\ x$
$\frac{du}{dx}\ =\ sec\ x\ tan\ x$
$du\ =\ sec\ x\ tan\ x\ dx$
$\int\ sec^6\ x\ sec\ x\ tan\ x\ dx = \int u^6\ du$
$=\frac{u^7}{7} + c$
$=\frac{sec^7\ x}{7} + c$
$\boxed{\int\ sec^7\ x\ tan\ x\ dx\ =\ \frac{sec^7\ x}{7} + c}$

$\color {purple} {Example\ 18\ .}\ \int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$put\ u\ =\ ax^2 + bx + c$
$\frac{du}{dx}= \ 2ax + b$

du  = (2ax + b) dx

$\int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx= \int\ \frac{du}{\sqrt{u}}$
$=\frac{u^{\frac{-1}{2} + 1}}{\frac{-1}{2} + 1} + c$
$=\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c$
$=2u^{\frac{1}{2}} + c$
$=2(ax^2+bx+c)^{\frac{1}{2}}\ +\ c$
$\boxed{\int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx\ =\ 2\ \sqrt{(ax^2 + bx + c)}\ + c}$

$\color {purple} {Example\ 19\ .}\ \color {red} {Evaluate:}\ \int\frac{(sin^{-1}\ x)^4}{\sqrt{1\ -\ x^2}}\ dx\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$put\ u\ =\sin^{-1}\ x$
$\frac{du}{dx}= \ \frac{1}{\sqrt{1\ -\ x^2}}$
$du\ =\ \frac{1}{\sqrt{1\ -\ x^2}}\ dx$
$\int\frac{(sin^{-1}\ x)^4}{\sqrt{1\ -\ x^2}}\ dx\ = \int\ u^4\ du$
$=\ \frac{u^5}{5} + c$
$=\ \frac{(sin^{-1}\ x)^5}{5}\ +\ c$
$\boxed{\int\frac{(sin^{-1}\ x)^4}{\sqrt{1\ -\ x^2}}\ dx\ =\ \frac{(sin^{-1}\ x)^5}{5}\ + c}$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \color {red} {Evaluate\ :}\ \int\ sec^2\ 5\ x\ dx\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {2\ .}\ \color {red} {Evaluate\ :}\ \int\ cos^3\ 7\ x\ dx\ \hspace{15cm}$
$\color {purple} {3\ .}\ \color {red} {Evaluate\ :}\ \int\ \frac{2x\ -\ 1}{{\sqrt{(x^2\ -\ x\ -\ 1)}}}\ dx\ \hspace{15cm}$
$\color {purple} {4\ .}\ \color {red} {Evaluate\ :}\ \int\frac{sec^2\ x}{5\ +\ 4\ tan\ x}\ dx\ \hspace{15cm}$
$\color {purple} {5\ .}\ \color {red} {Evaluate\ :}\ \int\frac{(tan^{-1}\ x)^3}{1\ +\ x^2}\ dx\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {6\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{x\ +\ 1}{ x^2\ +\ 2x\ -\ 1}\ dx\ \hspace{2cm}\ (ii)\ \int\frac{sec^2\ x}{5\ +\ tan\ x}\ dx\ \hspace{10cm}$
$\color {purple} {7\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{6x^2\ -\ 1}{ 2x^3\ -\ x\ +\ 5}\ dx\ \hspace{2cm}\ (ii)\ \int\frac{sin\ \sqrt{x}}{\sqrt{x}}\ dx\ \hspace{10cm}$
$\color {purple} {8\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{cos\ x}{ (3\ -\ 5\ sin\ x)^6}\ dx\ \hspace{2cm}\ (ii)\ \int\frac{e^{tan^{-1}\ x}}{1\ +\ x^2}\ dx\ \hspace{10cm}$
$\color {purple} {9\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{e^x}{1\ +\ e^x}\ dx\ \hspace{2cm}\ (ii)\ \int\ tan^5\ x\ sec^2\ x\ dx\ \hspace{10cm}$