# METHODS OF INTEGRATION – INTEGRATION BY SUBSTITUTION (Excercise Problems with Solutions)

$\underline{PART\ -\ A}$
$1.\ \color {red}{Evaluate\ :\ \int\ sec^2\ 5\ x\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u\ =\ 5\ x$
$\frac{du}{dx}\ =\ 5$
$dx\ = \frac{1}{5}\ du$
$\int\ sec^2\ 5\ x\ dx = \int sec^2\ u\ du$
$=\ tan\ u\ +\ c$
$=\ tan\ 5\ x\ +\ c$
$\boxed{\int\ sec^2\ 5\ x\ dx\ =\ tan\ 5\ x\ + c}$
$\underline{PART\ -\ B}$
$2.\ \color {red}{Evaluate\ :\ \int\ cos^3\ 7\ x\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u\ =\ 7\ x$
$\frac{du}{dx}\ =\ 7$
$dx\ = \frac{1}{7}\ du$
$\int\ cos^3\ 7\ x\ dx\ =\ \frac{1}{7}\int cos^3\ u\ du$
$cos\ 3\ x\ =\ 4\ cos^3\ x\ -\ 3\ cos\ x$
$cos^3\ x\ =\ \frac{1}{4}[cos\ 3\ x\ +\ 3\ cos\ x]$
$cos^3\ u\ =\ \frac{1}{4}[cos\ 3\ u\ +\ 3\ cos\ u]$
$\int\ cos^3\ 7\ x\ dx\ =\ \frac{1}{7}\ \int\ \frac{1}{4}[cos\ 3\ u\ +\ 3\ cos\ u]\ dx$
$=\ \frac{1}{28}[\int cos\ 3\ u\ dx\ +\ 3\int\ cos\ u\ dx]$
$=\ \frac{1}{28}[\frac{sin\ 3\ u} {3}\ +\ 3\ sin\ u]\ + c$
$=\ \frac{1}{28}[\frac{sin\ 3\ (7\ x)} {3}\ +\ 3\ sin\ 7\ x]\ + c$
$=\ \frac{1}{28}[\frac{sin\ 21\ x} {3}\ +\ 3\ sin\ 7\ x]\ + c$
$\boxed{\int\ cos^3\ 7\ x\ dx\ =\ \frac{1}{28}[\frac{sin\ 21\ x} {3}\ +\ 3\ sin\ 7\ x]\ + c}$
$3.\ \color {red}{Evaluate\ :\ \int\ \frac{2x\ -\ 1}{{\sqrt{(x^2\ -\ x\ -\ 1)}}}\ dx}\ \hspace{15cm}$

Soln:

$put\ u\ =\ x^2\ -\ x\ -\ 1$
$\frac{du}{dx}= \ 2x\ -\ 1$
$du\ =\ (2\ x \ -\ 1)\ dx$
$\int\ \frac{2x\ -\ 1}{{\sqrt{(x^2\ -\ x\ -\ 1)}}}\ dx= \int\ \frac{du}{u}$
$=\ \int\ \frac{du}{\sqrt{u}}$
$=\frac{u^{\frac{-1}{2} + 1}}{\frac{-1}{2} + 1} + c$
$=\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c$
$=2u^{\frac{1}{2}} + c$
$=2(x^2\ -\ x\ -\ 1)^{\frac{1}{2}}$
$\boxed{\int\ \frac{2x\ -\ 1}{{\sqrt{(x^2\ -\ x\ -\ 1)}}}\ dx\ =\ 2\ \frac{1}{{\sqrt{(x^2\ -\ x\ -\ 1)}}} + c}$
$4.\ \color{red}{Evaluate:\ \int\frac{sec^2\ x}{5\ +\ 4\ tan\ x}\ dx}\ \hspace{15cm}$

Soln:

$put\ u\ =\ 5\ +\ 4\ tan\ x$
$\frac{du}{dx}= \ 4\ sec^2\ x$
$\frac{1}{4}\ du\ =\ sec^2\ x\ dx$
$\int\frac{sec^2\ x}{5\ +\ 4\ tan\ x}\ dx\ =\ \frac{1}{4} \int\ \frac{du}{u}$
$=\ \frac{1}{4}\ log\ u\ + c$
$=\ \frac{1}{4}\ log(5\ +\ 4\ tan\ x)\ + c$
$\boxed{\int\ \frac{sec^2\ x}{5\ +\ 4\ tan\ x}\ dx\ =\ \frac{1}{4}\ log(5\ +\ 4\ tan\ x)\ + c}$
$5.\ \color{red}{Evaluate:\ \int\frac{(tan^{-1}\ x)^3}{1\ +\ x^2}\ dx}\ \hspace{15cm}$

Soln:

$put\ u\ =\tan^{-1}\ x$
$\frac{du}{dx}= \ \frac{1}{1\ +\ x^2}$
$du\ =\ \frac{1}{1\ +\ x^2}\ dx$
$\int \frac{(tan^{-1}\ x)^3}{1\ +\ x^2}\ dx\ = \int\ u^3\ du$
$=\ \frac{u^4}{4} + c$
$=\ \frac{(tan^{-1}\ x)^4}{4}\ +\ c$
$\boxed{\int\ \frac{(tan^{-1}\ x)^3}{1\ +\ x^2}\ dx\ =\ \frac{(tan^{-1}\ x)^4}{4}\ + c}$
$\underline{PART\ -\ C}$
$6.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ \frac{x\ +\ 1}{ x^2\ +\ 2x\ -\ 1}\ dx\ \hspace{2cm}\ (ii)\ \int\frac{sec^2\ x}{5\ +\ tan\ x}\ dx}\ \hspace{10cm}$

Soln:

$(i)\ \int\ \frac{x\ +\ 1}{ x^2\ +\ 2x\ -\ 1}\ dx\ \hspace{10cm}$
$put\ u\ =\ x^2\ +\ 2x\ -\ 1$
$\frac{du}{dx}= \ 2x + 2$
$du\ =\ 2(x\ +\ 1)\ dx$
$\frac{1}{2}\ du\ =\ (x\ +\ 1)\ dx$
$\int\ \frac{x+ 1}{x^2\ +\ 2x\ -\ 1}\ dx\ =\ \frac{1}{2}\int\ \frac{du}{u}$
$=\ \frac{1}{2}\ log\ u + c$
$=\ \frac{1}{2}\ log( x^2\ +\ 2x\ -\ 1) + c$
$\boxed{\int\ \frac{x+ 1}{x^2\ +\ 2x\ -\ 1}\ dx\ =\ \frac{1}{2}\ log( x^2\ +\ 2x\ -\ 1)\ + c}$
$(ii)\ \int\ \frac{sec^2\ x}{5\ +\ tan\ x}\ dx\ \hspace{10cm}$
$put\ u\ =\ 5\ +\ tan\ x$
$\frac{du}{dx}= \ sec^2\ x$
$du\ =\ sec^2\ x\ dx$
$\int\frac{sec^2\ x}{5\ +\ tan\ x}\ dx\ =\ \int\ \frac{du}{u}$
$=\ log\ u\ + c$
$=\ log(5\ +\ tan\ x)\ + c$
$\boxed{\int\ \frac{sec^2\ x}{5\ +\ tan\ x}\ dx\ =\ log(5\ +\ 4\ tan\ x)\ + c}$
$7.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ \frac{6x^2\ -\ 1}{ 2x^3\ -\ x\ +\ 5}\ dx\ \hspace{2cm}\ (ii)\ \int\frac{sin\ \sqrt{x}}{\sqrt{x}}\ dx}\ \hspace{10cm}$

Soln:

$(i)\ \int\ \frac{6x^2\ -\ 1}{ 2x^3\ -\ x\ +\ 5}\ dx\ \hspace{10cm}$
$put\ u\ =\ 2x^3\ -\ x\ +\ 5$
$\frac{du}{dx}= \ 6x^2\ -\ 1$
$du\ =\ (6x^2\ -\ 1)\ dx$
$\int\ \frac{6x^2\ -\ 1}{ 2x^3\ -\ x\ +\ 5}\ dx\ =\ \int\ \frac{du}{u}$
$=\ log\ u\ + c$
$=\ log(2x^3\ -\ x\ +\ 5)\ + c$
$\boxed{\int\ \frac{6x^2\ -\ 1}{ 2x^3\ -\ x\ +\ 5}\ dx\ =\ log(2x^3\ -\ x\ +\ 5)\ + c}$
$(ii)\ \int\ \frac{sin\ \sqrt{x}}{\sqrt{x}}\ dx\ \hspace{10cm}$
$put\ u\ =\ \sqrt{x}$
$\frac{du}{dx}= \ \frac{1}{2\sqrt{x}}$
$2\ du\ =\ \frac{1}{\sqrt{x}}\ dx$
$\int\ \frac{sin\ \sqrt{x}}{\sqrt{x}}\ dx\ =\ \int\ 2\ sin\ u\ du$
$=\ -\ 2\ cos\ u\ + c$
$=\ -\ 2\ cos\ \sqrt{x}\ + c$
$\boxed{\int\ \frac{sin\ \sqrt{x}}{\sqrt{x}}\ dx\ =\ -\ 2\ cos\ \sqrt{x}\ + c}$
$8.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ \frac{cos\ x}{ (3\ -\ 5\ sin\ x)^6}\ dx\ \hspace{2cm}\ (ii)\ \int\frac{e^{tan^{-1}\ x}}{1\ +\ x^2}\ dx}\ \hspace{10cm}$

Soln:

$(i)\ \int\ \frac{cos\ x}{ (3\ -\ 5\ sin\ x)^6}\ dx\ \hspace{10cm}$
$put\ u\ =\ 3\ -\ 5\ sin\ x$
$\frac{du}{dx}= \ -\ 5\ cos\ x$
$\frac{-1}{5}du\ =\ cos\ x\ dx$
$\int\ \frac{cos\ x}{ (3\ -\ 5\ sin\ x)^6}\ dx\ =\ \frac{-1}{5}\int\ \frac{du}{u^6}$
$=\ \frac{-1}{5} \int\ u^{-6}\ du$
$=\frac{-1}{5} (\frac{u^{-5}}{-5}) + c$
$=\frac{(3\ -\ 5\ sin\ x)^{-5}}{25} + c$
$\boxed{\int\ \frac{cos\ x}{ (3\ -\ 5\ sin\ x)^6}\ dx\ =\ \frac{(3\ -\ 5\ sin\ x)^{-5}}{25}\ + c}$
$(ii)\ \int\ \frac{e^{tan^{-1}\ x}}{1\ +\ x^2}\ dx\ \hspace{10cm}$
$put\ u\ =\tan^{-1}\ x$
$\frac{du}{dx}= \ \frac{1}{1\ +\ x^2}$
$du\ =\ \frac{1}{1\ +\ x^2}\ dx$
$\int \frac{e^{tan^{-1}\ x}}{1\ +\ x^2}\ dx\ = \int\ e^u\ du$
$=\ e^u + c$
$=\ e^{tan^{-1}\ x} + c$
$\boxed{\int\ \frac{e^{tan^{-1}\ x}}{1\ +\ x^2}\ dx\ =\ e^{tan^{-1}\ x}\ + c}$
$9.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ \frac{e^x}{1\ +\ e^x}\ dx\ \hspace{2cm}\ (ii)\ \int\ tan^5\ x\ sec^2\ x\ dx}\ \hspace{10cm}$

Soln:

$(i)\ \int\ \frac{e^x}{1\ +\ e^x}\ dx\ \hspace{10cm}$
$put\ u\ =\ 1\ +\ e^x$
$\frac{du}{dx}= \ e^x$
$du\ =\ e^x\ dx$
$\int\ \frac{e^x}{1\ +\ e^x}\ dx= \int\ \frac{du}{u}$
$=\ log\ u\ + c$
$=\ log(1\ +\ e^x)\ + c$
$\boxed{\int\ \frac{e^x}{1\ +\ e^x}\ dx\ =\ log(1\ +\ e^x)\ + c}$
$(ii)\ \int\ tan^5\ x\ sec^2\ x\ dx\ \hspace{10cm}$
$put\ u\ =\ tan\ x$
$\frac{du}{dx}\ =\ sec^2\ x$
$du\ =\ sec^2\ x\ dx$
$\int\ tan^5\ x\ sec^2\ x\ dx = \int u^5\ du$
$=\frac{u^6}{6} + c$
$=\frac{tan^6 x}{6} + c$
$\boxed{\int\ tan^5\ x\ sec^2\ x\ dx\ =\ \frac{tan^6 x}{6} + c}$