Examples On Integration By Parts Set-2

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Example - 43

Evaluate the following integrals

(a) \(\int {\sqrt {{a^2} + {x^2}} dx} \) (b) \(\int {\sqrt {{a^2} - {x^2}} dx} \) (c) \(\int {\sqrt {{x^2} - {a^2}} dx} \)

Solution: In all these three questions, we use integration by parts, taking unity as the second function:

(a)                                                                                       \(I = \int {\mathop {\sqrt {{a^2} + {x^2}} }\limits_{{\rm{Ist}}} \cdot \mathop 1\limits_{{\rm{IInd}}} \,\,dx} \)

\[\begin{align}&\qquad\qquad = x\sqrt {{a^2} + {x^2}} - \int {\frac{{{x^2}}}{{\sqrt {{a^2} + {x^2}} }}\,\,dx} \\&\qquad\qquad = x\sqrt {{a^2} + {x^2}} - \int {\frac{{({a^2} + {x^2}) - {a^2}}}{{\sqrt {{a^2} + {x^2}} }}\,\,dx} \\&\qquad\qquad = x\sqrt {{a^2} + {x^2}} - \int {\sqrt {{a^2} + {x^2}} \,\,dx} + {a^2}\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} \,dx\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \mathop \uparrow \limits_{\scriptstyle{\rm{of the standard }}\atop\scriptstyle{\rm{form (22)}}} \\&\qquad\qquad = x\sqrt {{a^2} + {x^2}} - I + {a^2}\ln \left| {x + \sqrt {{x^2} + {a^2}} } \right|\\ & \Rightarrow\qquad I = \frac{1}{2}\left\{ {x\sqrt {{a^2} + {x^2}} + {a^2}\ln \left| {x + \sqrt {{x^2} + {a^2}} } \right|} \right\} + C\end{align}\]

 

(b)                                                                                     \(I = \int {\mathop {\sqrt {{a^2} - {x^2}} }\limits_{{\rm{Ist}}} \cdot \mathop 1\limits_{{\rm{IInd}}} \,\,dx} \)

\[\begin{align} &\qquad\qquad = x\sqrt {{a^2} - {x^2}} + \int {\frac{{{x^2}}}{{\sqrt {{a^2} - {x^2}} }}\,\,dx} \\&\qquad\qquad = x\sqrt {{a^2} - {x^2}} - \int {\frac{{({a^2} - {x^2}) - {a^2}}}{{\sqrt {{a^2} - {x^2}} }}\,\,dx} \\&\qquad\qquad = x\sqrt {{a^2} - {x^2}} - \int {\sqrt {{a^2} - {x^2}} \,\,dx + {a^2}\int {\frac{1}{{\sqrt {{a^2} - {x^2}} }}} \,\,dx} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\quad\quad\qquad \mathop \uparrow \limits_{\scriptstyle{\rm{Of the standard }}\atop\scriptstyle{\rm{form (15)}}} \\&\qquad\qquad = x\sqrt {{a^2} - {x^2}} - I + {a^2}{\sin ^{ - 1}}\frac{x}{a}\\ &\Rightarrow\qquad I = \frac{1}{2}\left\{ {x\sqrt {{a^2} - {x^2}} + {a^2}{{\sin }^{ - 1}}\frac{x}{a}} \right\} + C\end{align}\]

 

(c)                                                                                  \(I = \int {\mathop {\sqrt {{x^2} - {a^2}} }\limits_{{\rm{Ist}}} \cdot \mathop 1\limits_{{\rm{IInd}}} \,\,dx} \)

\[\begin{align}&\qquad\qquad = x\sqrt {{x^2} - {a^2}} - \int {\frac{{{x^2}}}{{\sqrt {{x^2} - {a^2}} }}\,\,dx} \\&\qquad\qquad = x\sqrt {{x^2} - {a^2}} - \int {\frac{{({x^2} - {a^2}) + {a^2}}}{{\sqrt {{x^2} - {a^2}} }}\,\,dx} \\&\qquad\qquad = x\sqrt {{x^2} - {a^2}} - \int {\sqrt {{x^2} - {a^2}} \,\,dx - {a^2}\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}\,\,dx} } \\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \mathop \uparrow \limits_{\scriptstyle{\rm{Of the standard}}\atop\scriptstyle{\rm{form (23)}}} \\&\qquad\qquad= x\sqrt {{x^2} - {a^2}} - I - {a^2}\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right|\\& \Rightarrow\qquad I = \frac{1}{2}\left\{ {x\sqrt {{x^2} - {a^2}} - {a^2}\ln |x + \sqrt {{x^2} - {a^2}} } \right\} + C\end{align}\]

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These three results have an extensive use because we are now in a position to integrate expression of the form \(\int {\sqrt {Q(x)} \,\,dx} \) (where \(Q(x)\) is a quadratic polynomial) or other expressions reducible to this form.

Due to their wide use, we will soon extend Table-1: Standard Integrals to include these three

Example - 44

Evaluate \(\int {{e^{ax}}\sin bx\,\,dx\,\,{\rm{and }}\int {{e^{ax}}\cos bx\,\,dx} } \)

Solution: Let                                                            \({I_1} = \int {{e^{ax}}\sin bx\,\,dx\,\,{\rm{and}}\,\,{I_2} = \int {{e^{ax}}\cos bx\,\,dx} } \)

\[\begin{align}&\;\;\quad\quad {I_1} = \int {{{\mathop e\limits_{{\rm{Ist}}} }^{ax}}\mathop {\sin bx}\limits_{{\rm{ IInd}}} \,\,dx} \left( {Use\;{\rm{ }}integration\;{\rm{ }}by\;{\rm{ }}parts} \right)\\&\quad\qquad\;\; = \frac{{ - {e^{ax}}\cos bx}}{b} + \frac{a}{b}\int {{e^{ax}}\cos bx\,\,dx} \\ &\Rightarrow\quad {I_1} = \frac{{ - {e^{ax}}\cos bx}}{b} + \frac{a}{b}{I_2}\qquad\qquad\qquad...\left( 1 \right)\end{align}\]

Similarly, we now apply integration by parts on \({I_2}\) :

\[\begin{align} &\quad\quad\;\; {I_2} = \int {{{\mathop e\limits_{{\rm{Ist}}} }^{ax}}\mathop {\cos bx}\limits_{{\rm{IInd}}} \,\,dx} \\&\qquad\quad\;\; = \frac{{{e^{ax}}\sin bx}}{b} - \frac{a}{b}\int {{e^{ax}}\sin bx\,\,dx} \\&\Rightarrow\quad {I_2} = \frac{{{e^{ax}}\sin bx}}{b} - \frac{a}{b}{I_1}\qquad\qquad\qquad...\left( 2 \right)\end{align} \]

Solving (1) and (2), we have (verify):

\[\begin{align} {I_1} = \frac{{{e^{ax}}}}{{{a^2} + {b^2}}}(a\sin bx - b\cos bx) + C\\{I_2} = \frac{{{e^{ax}}}}{{{a^2} + {b^2}}}(a\cos bx + b\sin bx) + C\end{align}\]

We now include the results of this and the last example in our table of standard integrals:

Table - 1 Continued

Standard integrals (evaluated by parts)

Integral

Result

26. \(\int {\sqrt {{a^2} + {x^2}} \,\,dx} \) \(\begin{align}\frac{1}{2}\left\{ {x\sqrt {{a^2} + {x^2}} + {a^2}\ln \left| {x + \sqrt {{x^2} + {a^2}} } \right|} \right\} + C\end{align}\)
27. \(\int {\sqrt {{a^2} - {x^2}} \,\,dx} \) \(\begin{align}\frac{1}{2}\left\{ {x\sqrt {{a^2} - {x^2}} + {a^2}{{\sin }^{ - 1}}\frac{x}{a}} \right\} + C\end{align}\)
28. \(\int {\sqrt {{x^2} - {a^2}} \,\,dx} \) \(\begin{align}\frac{1}{2}\left\{ {x\sqrt {{x^2} - {a^2}} - {a^2}\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right|} \right\} + C\end{align}\)
29. \(\int {{e^{ax}}\sin bx\,\,dx} \) \(\begin{align}\frac{{{e^{ax}}}}{{{a^2} + {b^2}}}(a\sin bx - b\cos bx) + C\end{align}\)
30. \(\int {{e^{ax}}\cos bx\,\,dx} \) \(\begin{align}\frac{{{e^{ax}}}}{{{a^2} + {b^2}}}(a\cos bx + b\sin bx) + C\end{align}\)
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