Examples On Integration By Parts Set-3

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Example – 45

Evaluate the following integrals.

(a) \(\begin{align}\int {\sqrt {{x^2} + x + 1} \,\,dx} \end{align}\) (b) \(\begin{align}\int {\sqrt {1 - 2x - {x^2}} \,\,dx}\end{align} \)
(c) \(\begin{align}\int {\sqrt {{x^2} + 3x + 2} \,\,dx}\end{align} \) (d) \(\begin{align}\int {x\,\,{e^x}\cos x\,\,dx} \end{align}\)

Solution: (a) We can write \({x^2} + x + 1\) as

\[\begin{align} & {x^2} + x + 1 = \frac{3}{4} + {\left( {x + \frac{1}{2}} \right)^2}\\ &\qquad\qquad\;\;= {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( {x + \frac{1}{2}} \right)^2}\end{align}\]

Thus, this integral is of the standard form (26):

\[I = \frac{1}{2}\left\{ {\left( {x + \frac{1}{2}} \right)\sqrt {{x^2} + x + 1} + \frac{3}{4}\ln \left| {\left( {x + \frac{1}{2}} \right) + \sqrt {{x^2} + x + 1} } \right|} \right\} + C\]

(b) We can write \(1 - 2x - {x^2}\) upon rearrangement as

\[\begin{align} 1 - 2x - {x^2} = 2 - (1 + 2x + {x^2})\\ = {\left( {\sqrt 2 } \right)^2} - {\left( {x + 1} \right)^2}\end{align}\]


This integral therefore of  the standard form (27) :

\[I = \frac{1}{2}\left\{ {(x + 1)\sqrt {1 - 2x - {x^2}} + 2{{\sin }^{ - 1}}\left( {\frac{{x + 1}}{{\sqrt 2 }}} \right)} \right\} + C\]

(c) Upon rearrangement, \({x^2} + 3x + 2\) can be written as

\[{x^2} + 3x + 2 = {\left( {x + \frac{3}{2}} \right)^2} - \frac{1}{4}\\ \qquad\qquad\qquad\quad = {\left( {x + \frac{3}{2}} \right)^2} - {\left( {\frac{1}{2}} \right)^2}\]

Thus, this integral is of the standard form (28):

\[I = \frac{1}{2}\left\{ {\left( {x + \frac{3}{2}} \right)\sqrt {{x^2} + 3x + 2} - \frac{1}{4}\ln \left| {\left( {x + \frac{3}{2}} \right) + \sqrt {{x^2} + 3x + 2} } \right.} \right\} + C\]

(d) From the standard form (30), we know the integral of \({{e}^{x}}\cos x\) :

\[\int {{e^x}\cos x\,\,dx = \frac{1}{2}{e^x}\left( {\sin x + \cos x} \right)} + C\]

Thus, we can apply integration by parts on the expression \(x\,{{e}^{x}}\cos x,taking\;x\), taking as the first function:


\[\begin{align}& I = \int {\mathop x\limits_{{\rm{Ist}}} ({e^x}\mathop {\cos x}\limits_{{\rm{IInd}}} )dx} \\  &= \frac{1}{2}x{e^x}(\sin x + \cos x) - \int {1 \cdot \left( {\frac{1}{2}{e^x}(\sin x + \cos x)} \right)dx} \\& = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{1}{2}\left\{ {{e^x}\sin x\,\,dx + \int {{e^x}\cos x\,\,dx} } \right\}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\mathop \nearrow \limits_{{\rm{standard form (29)}}} \mathop \nearrow \limits_{{\rm{standard form (30)}}} \\&  = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{{{e^x}}}{4}\left\{ {(\sin x - \cos x) + (\cos x + \sin x)} \right\} + C\\& = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{1}{2}{e^x}\sin x + C\end{align}\]
 

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We now consider an integral of the form \(\int {L(x)\sqrt {Q(x)} \,\,dx} \) where \(L(x)\) is a linear factor and \(Q(x)\) is a quadratic factor. As described earlier, we can always find constants \(\alpha \,\,{\rm{and }}\beta \) such that

\[L(x) = \alpha Q'(x) + \beta \]

so that

\[\begin{align}& I = \int {L(x)\sqrt {Q(x)} \,\,dx}\\&= \alpha \int {Q'(x)\sqrt {Q(x)} \,\,dx + \beta \int {\sqrt {Q(x)} \,\,dx} } \\& = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{I_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;{I_2}\end{align}\]

\({I_1}\) can be solved by the substitution \(Q(x) = t\) so that \(Q'(x)dx = dt\) and \({I_1}\) reduces to \(\alpha \int {\sqrt t \,\,dt} \) whereas \({I_2}\) is of the standard form (26), (27) or (28) depending on the form of \(Q(x)\) .

Example – 46

Evaluate the following integrals:

(a) \(\int {(x + 3)\sqrt {{x^2} + 2x + 3} } \,\,dx\) (b) \(\begin{align}\int {\frac{x}{{x - \sqrt {{x^2} - 1} }}} \,\,dx\end{align}\)

Solution : (a) The derivative of \({x^2} + 2x + 3 \;is \;2x + 3\) . Thus, we can find \(\alpha \,\,{\rm{and }}\beta \) such that

\[\begin{align}&\qquad (x + 3) = \alpha (2x + 3) + \beta\\& \Rightarrow\quad \alpha =\frac{1}{2},\,\,\,\beta =\frac{3}{2} \\& \Rightarrow\quad I = \frac{1}{2}\int {(2x + 3)\sqrt {{x^2} + 2x + 3} } \,\,dx + \frac{3}{2}\int {\sqrt {{x^2} + 2x + 3} } \,\,dx\\ &\qquad\quad = \mathop {\frac{1}{2}\int {\sqrt t } \,\,dt\,\,\,\,\,\,}\limits_{({\text{where }}t{\text{ = }}{x^{\rm{2}}} + 2x + 3)} + \mathop {\frac{3}{2}\int {\sqrt {{{(x + 1)}^2} + 2} } \,\,dx}\limits_{({\text{of the standard form (26)}})} \\&\qquad\quad = \frac{{{t^{3/2}}}}{3} + \frac{3}{4}\left\{ {(x + 1)\sqrt {{x^2} + 2x + 3} + 2\ln \left| {(x + 1) + \sqrt {{x^2} + 2x + 3} } \right|} \right\} + C\\ &\qquad\quad= \frac{{{{({x^2} + 2x + 3)}^{3/2}}}}{3} + \frac{3}{4}\left\{ {(x + 1)\sqrt {{x^2} + 2x + 3} + 2\ln \left| {(x + 1) + \sqrt {{x^2} + 2x + 3} } \right|} \right\} + C\end{align}\]

(b) We first rationalize the denominator:

\[\begin{align}& I = \int {\frac{x}{{x - \sqrt {{x^2} - 1} }}\,\,dx}\\& = \int {x(x + \sqrt {{x^2} - 1} )\,\,dx} \\& = \int {{x^2}dx + \mathop {\int {x\sqrt {{x^2} - 1} } \,\,dx}\limits_{{\text{(substitute }}{x^{\text{2}}}{\text{-1 = }}t{\text{)}}} } \\& = \frac{{{x^3}}}{3} + \frac{1}{2}\int {\sqrt t \,\,dt} \\& = \frac{{{x^3}}}{3} + \frac{{{t^{3/2}}}}{3} + C\\& = \frac{{{x^3}}}{3} + \frac{{{{({x^2} - 1)}^{3/2}}}}{3} + C\end{align}\]

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