# INTEGRATION BY PARTS (Excercise Problems with Solutions)

$\underline{PART\ -\ B}$
$1.\ \color {red}{Evaluate\ :\ \int x\ e^{-2x}\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$ILATE$
$u= x\ \hspace{2cm}\ dv = e^{-2x}\ dx$
$\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int e^{-2x}\ dx$
$\frac{du}{dx} = 1\ \hspace{2cm}\ v = \frac{e^{-2x}}{-2}$
$\ du = dx\ \hspace{5cm}$
$\int u\ dv = uv – \int v\ du$
$\int x\ e^{-2x} dx = x\ \frac{e^{-2x}}{-2} – \int \frac{e^{-2x}}{-2}\ dx$
$=-\ x\ \frac{e^{-2x}}{2} +\ \frac{1}{2} \frac{e^{-2x}}{-2}\ + c$
$\boxed{\int x\ e^{-2x} dx = -\ x\ \frac{e^{-2x}}{2} – \frac{e^{-2x}}{4} +c}$
$2.\ \color {red}{Evaluate\ :\ \int x\ sec^2 x\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$ILATE$
$u= x\ \hspace{2cm}\ dv = sec^2 x\ dx$
$\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int sec^2 x\ dx$
$\frac{du}{dx} = 1\ \hspace{2cm}\ v = tan\ x$
$\ du = dx\ \hspace{5cm}$
$\int u\ dv = uv – \int v\ du$
$\int x\ sec^2 x\ dx = x\ tan\ x – \int tan\ x\ dx$
$= x\ tan\ x\ +\ log\ sec\ x +c$
$\boxed{\int x\ sec^2 x\ dx = x\ tan\ x\ +\ log\ sec\ x +c}$
$\underline{PART\ -\ C}$
$3.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ x\ sin\ 5x\ dx\ \hspace{2cm}\ (ii)\ \int x\ log\ x\ dx}\ \hspace{10cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(i)\ \int\ x\ sin\ 5x\ dx\ \hspace{10cm}$
$ILATE$
$u= x\ \hspace{2cm}\ dv =sin\ 5x\ dx$
$\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int sin\ 5x\ dx$
$\frac{du}{dx} = 1\ \hspace{2cm}\ v = -\frac{cos\ 5x}{5}$
$\ du = dx\ \hspace{5cm}$
$\int u\ dv = uv – \int v\ du$
$\int x\ sin\ 5x\ dx = – x\ \frac{cos\ 5x}{5} + \int \frac{cos\ 5x}{5}\ dx$
$= – x\ \frac{cos\ 5x}{5} + \frac{1}{5}(\frac{sin\ 5x}{5})\ + c$
$= – x\ \frac{cos\ 5x}{5} +\frac{sin\ 5x}{25}\ + c$
$\boxed{\int x\ sin\ 5x\ dx = – x\ \frac{cos\ 5x}{5} +\frac{sin\ 5x}{25}\ + c}$
$(ii)\ \int\ x\ log\ x\ dx\ \hspace{10cm}$
$ILATE$
$u= log\ x\ \hspace{2cm}\ dv = x\ dx$
$\frac{du}{dx} = \frac{d}{dx} ( log\ x)\ \hspace{2cm}\ \int dv = \int x\ dx$
$\frac{du}{dx} = \frac{1}{x}\ \hspace{2cm}\ v = \frac{x^2}{2}$
$\ du = \frac{1}{x}\ dx\ \hspace{5cm}$
$\int u\ dv = uv – \int v\ du$
$\int x\ log\ x\ dx = log\ x\ \frac{x^2}{2} – \int \frac{x^2}{2}\ (\frac{1}{x})\ dx$
$= log\ x\ \frac{x^2}{2} – \frac{1}{2}\ \int x\ dx$
$= log\ x\ \frac{x^2}{2} – \frac{1}{2}\ \frac{x^2}{2}\ + c$
$= log\ x\ \frac{x^2}{2} – \frac{x^2}{4}\ + c$
$\boxed{\int x\ log\ x\ dx = log\ x\ \frac{x^2}{2} – \frac{x^2}{4}\ + c}$
$4.\ \color {red}{Evaluate:\ \int x\ cos\ 3x\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$ILATE$
$u= x\ \hspace{2cm}\ dv =cos\ 3x\ dx$
$\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int cos\ 3x\ dx$
$\frac{du}{dx} = 1\ \hspace{2cm}\ v = \frac{sin\ 3x}{3}$
$\int u\ dv = uv – \int v\ du$
$\int x\ cos\ 3x\ dx = x\ \frac{sin\ 3x}{3} – \int \frac{sin\ 3x}{3}\ dx$
$= x\ \frac{sin\ 3x}{3} – \frac{1}{3}(\frac{cos\ 3x}{3})\ + c$
$= x\ \frac{sin\ 3x}{3} +\frac{cos\ 3x}{9}\ + c$
$\boxed{\int x\ cos\ 3x\ dx = x\ \frac{sin\ 3x}{3} +\frac{cos\ 3x}{9}\ + c}$
$5.\ \color {red}{Evaluate:\ \int x\ e^{4x}\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$ILATE$
$u= x\ \hspace{2cm}\ dv = e^{4x}\ dx$
$\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int e^{4x}\ dx$
$\frac{du}{dx} = 1\ \hspace{2cm}\ v = \frac{e^{4x}}{4}$
$\ du = dx\ \hspace{5cm}$
$\int u\ dv = uv – \int v\ du$
$\int x\ e^{4x} dx = x\ \frac{e^{4x}}{4} – \int \frac{e^{4x}}{4}\ dx$
$= x\ \frac{e^{4x}}{4} – \frac{e^{4x}}{16}\ + c$
$\boxed{\int x\ e^{4x}\ dx = x\ \frac{e^{4x}}{4} – \frac{e^{4x}}{16} +c}$
$6.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ x^3\ log\ x\ dx\ \hspace{2cm}\ (ii)\ \int x\ e^{-5x}\ dx}\ \hspace{10cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(i)\ \int\ x^3\ log\ x\ dx\ \hspace{10cm}$
$ILATE$
$u= log\ x\ \hspace{2cm}\ dv = x^3\ dx$
$\frac{du}{dx} = \frac{d}{dx} ( log\ x)\ \hspace{2cm}\ \int dv = \int x^3\ dx$
$\frac{du}{dx} = \frac{1}{x}\ \hspace{2cm}\ v = \frac{x^4}{4}$
$\ du = \frac{1}{x}\ dx\ \hspace{5cm}$
$\int u\ dv = uv – \int v\ du$
$\int x^3\ log\ x\ dx = log\ x\ \frac{x^4}{4} – \int \frac{x^4}{4}\ (\frac{1}{x})\ dx$
$= log\ x\ \frac{x^4}{4} – \frac{1}{4}\ \int x^3\ dx$
$= log\ x\ \frac{x^4}{4} – \frac{1}{4}\ \frac{x^4}{4}\ + c$
$= log\ x\ \frac{x^4}{4} – \frac{x^4}{16}\ + c$
$\boxed{\int x^3\ log\ x\ dx = log\ x\ \frac{x^4}{4} – \frac{x^4}{16}\ + c}$
$(ii)\ \int\ x\ e^{-5x}\ dx\ \hspace{10cm}$
$ILATE$
$u= x\ \hspace{2cm}\ dv = e^{-5x}\ dx$
$\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int e^{-5x}\ dx$
$\frac{du}{dx} = 1\ \hspace{2cm}\ v = \frac{e^{-5x}}{-5}$
$\ du = dx\ \hspace{5cm}$
$\int u\ dv = uv – \int v\ du$
$\int x\ e^{-5x} dx = x\ \frac{e^{-5x}}{-5} – \int \frac{e^{-5x}}{-5}\ dx$
$=-\ x\ \frac{e^{-5x}}{5} + \frac{1}{5} \frac{e^{-5x}}{-5}\ + c$
$\boxed{\int x\ e^{-5x} dx = -\ x\ \frac{e^{-5x}}{5} – \frac{e^{-5x}}{25} +c}$