Examples On Integration By Parts Set-4

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We now consider some miscenalleous examples of integration by parts:

Example – 47

Evaluate the following integrals

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

Solution: (a) We apply integration by parts, taking unity as the second function.

\[\begin{align}& I = \int {\mathop {{{\tan }^{ - 1}}}\limits_{{\rm{Ist}}} \sqrt x } \cdot \mathop {1\,\,}\limits_{{\rm{IInd}}} dx\\ &= x{\tan ^{ - 1}}\sqrt x - \int {\frac{1}{{1 + {{(\sqrt x )}^2}}} \cdot \frac{1}{{2\sqrt x }} \cdot x\,\,dx} \\& = x{\tan ^{ - 1}}\sqrt x - \frac{1}{2}\int {\frac{{\sqrt x }}{{1 + x}}\,\,dx} \end{align}\]

To evaluate the integral on the right side above (call it \({I_2}\) ), we let \(x = {t^2}\) . Thus,


\[\begin{align}& dx = 2tdt\\& {I_2} = \int {\frac{{{t^2}}}{{1 + {t^2}}}\,\,dt}\\ &\quad = \int {\left( {1 - \frac{1}{{1 + {t^2}}}} \right)\,\,dt} \\&\quad = t - {\tan ^{ - 1}}t + C\\&\quad= \sqrt x - {\tan ^{ - 1}}\sqrt x + C\end{align}\]

Thus,                                                    \(I = x{\tan ^{ - 1}}\sqrt x + \sqrt x - {\tan ^{ - 1}}\sqrt x + C\)

(b) We can divide the given expression into two parts, \(\begin{align}\frac{x}{{\sqrt {1 + {x^2}} }}{\rm{ \;and \;}}{\tan ^{ - 1}}x\end{align}\) . The integration of the first expression is easy to carry out:

                                                                                                                \(\begin{align}\int {\frac{x}{{\sqrt {1 + {x^2}} }}} \,\,dx = \frac{1}{2}\int {\frac{{dt}}{{\sqrt t }}} \left( where\;t=\text{ }1\text{ }+{{x}^{2}} \right) \end{align}\) 

\[\begin{align}&= \sqrt t \\& = \sqrt {1 + {x^2}} \end{align}\]

Now we integrate our original expression by parts:


\[\begin{align}& I = \int {{{\mathop {\tan }\limits_{{\rm{Ist}}} }^{ - 1}}x \cdot \frac{x}{{\mathop {\sqrt {1 + {x^2}} }\limits_{{\rm{IInd}}} }}dx} \\& = \sqrt {1 + {x^2}} {\tan ^{ - 1}}x - \int {\frac{1}{{1 + {x^2}}} \cdot \sqrt {1 + {x^2}} } \,\,dx\\& = \sqrt {1 + {x^2}} {\tan ^{ - 1}}x - \mathop {\int {\frac{1}{{\sqrt {1 + {x^2}} }}} \,\,dx}\limits_{{\text{(of the standard form (22))}}} \\& = \sqrt {1 + {x^2}} {\tan ^{ - 1}}x - \ln \left| {x + \sqrt {1 + {x^2}} } \right| + C\end{align}\]

Example - 48

Evaluate \(\begin{align}\int {\frac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} \,\,dx\end{align}\)

Solution: Consider the expression \(\begin{align}\frac{1}{{x\sin x + \cos x}}\end{align}\) carefully. See what happens when we differentiate this expression:

\[{\left( {\frac{1}{{x\sin x + \cos x}}} \right)^′} = \frac{{ - x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}\]

Thus, from this relation, we can say that

\[\int {\frac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dx} = \frac{{ - 1}}{{x\sin x + \cos x}} + C\qquad\qquad...{\text { }}\left( 1 \right)\]

(1) gives us the required clue as to how to solve our original integral; we need to split the numerator \(x{{~}^{2}}\) of the original integral into two parts, \(\begin{align} x\cos x\,\,{\rm{and}}\frac{x}{{\cos x}}:\end{align}\)

\[I=\int{\frac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx=}\int{\underset{\begin{smallmatrix}  \\ 
 \text{Ist} \end{smallmatrix}}{\mathop{\frac{x}{\cos x}}}\,\,\,\cdot \underset{\begin{smallmatrix} 
  \\  \left( \begin{smallmatrix}  \text{We have evaluated the integral } \\  \text{of this IInd function in}\left( 1 \right) \end{smallmatrix} \right) \end{smallmatrix}}{\mathop{\frac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}}}\,}\,\,dx\]

\[\begin{align}& = \frac{{ - x}}{{\cos x\left( {x\sin x + \cos x} \right)}} - \int {\frac{{\cos x + x\sin x}}{{{{\cos }^2}x}}} \cdot \frac{{ - 1}}{{x\sin x + \cos x}} \cdot dx\\& = \frac{{ - x}}{{\cos x\left( {x\sin x + \cos x} \right)}} + \int {{{\sec }^2}x\,dx} \\& = \frac{{ - x}}{{\cos x\left( {x\sin x + \cos x} \right)}} + \tan x + C\end{align}\]

How to think of such non-trivial manipulations is the skill that you absolutely require if you want to master integration!

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