Question 1:
Obtain an integral of u sin u.
Answer 1:
Suppose, I = \(\int u\; sin\; u \;du\)
Integrating the equation by parts by taking u as first function and sin u as second function, we get,
\(I = u\;\int sin\; u \;du – \int \left \{ \left ( \frac{d}{du} \;u \right ) \int sin\; u \;du \right \} \;du \\ = u (- cos\; u) – \int 1. (- cos\; u) \;du = – u\; cos\; u + sin\; u + C\)
Question 2:
Obtain an integral of u sin 3u.
Answer 2:
Suppose, I = \(\int u\; sin\; u\; du\)
Integrating the equation by parts by taking u as first function and sin 3u as second function, we get,
\(I = u\;\int sin\; 3u \;du – \int \left \{ \left ( \frac{d}{du} \;u \right ) \int sin\; 3u \;du \right \} \;du \\ = u (\frac{- cos\; 3u}{3}) – \int 1. (\frac{- cos\; 3u}{3}) \;du \\ = \frac{- u\; cos\; 3u}{3} + \frac{1}{9} sin\; 3u + C\)
Question 3:
Obtain an integral of \(u^{2}. e^{u}\)
Answer 3:
Suppose, I = \(\int u^{2}. e^{u} \;du\)
Integrating the equation by parts by taking u2 as first function and eu as second function, we get,
\(I = u^{2} \int e^{u} \;du – \int \left \{ \left ( \frac{d}{du} u^{2} \right ) \int e^{u} \;du \right \} \;du \\ = u^{2} e^{u} – \int 2u\; e^{u} \;du \\ = u^{2} e^{u} – 2 \int u\; e^{u} \;du \\ Integrating\; by\; parts\;, we\; get, \\ = u^{2} e^{u} – 2 \left [u\; \int e^{u} \;du – \int \left \{ \left ( \frac{d}{du}\; u \right ). \int e^{u} \;du \right \} \right ] \;du \\ = u^{2} e^{u} – 2 \left [u\; e^{u} – \int e^{u} \;du \right ] \\ = u^{2} e^{u} – 2 \left [u\; e^{u} – e^{u} \right ] \\ = u^{2} e^{u} – 2 u\; e^{u} – 2 e^{u} + C \\ = e^{u} (u^{2} – 2u + 2) + C\)
Question 4:
Obtain an integral of u log u.
Answer 4:
Suppose, I = \(\int u\; log\; u \;du\)
Integrating the equation by parts by taking log u as first function and u as second function, we get,
\(I = log\; u \int u\; du – \int \left \{ \left ( \frac{d}{du} log\; u \right ) \int u \;du \right \} \;du \\ = log\; u. \frac{u^{2}}{2} – \int \frac{1}{u}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} log\; u}{2} – \int \frac{u}{2} \;du \\ = \frac{u^{2} log\; u}{2} = \frac{u^{2}}{4} + C\)
Question 5:
Obtain an integral of u log 2u.
Answer 5:
Suppose, I = \(\int u\; log\; 2u \;du\)
Integrating the equation by parts by taking log 2 u as first function and u as second function, we get,
\(I = log\; 2u \int u\; du – \int \left \{ \left ( \frac{d}{du} 2 log\; u \right ) \int u \;du \right \} \;du \\ = log\; 2u. \frac{u^{2}}{2} – \int \frac{2}{2 u}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} log\; 2u}{2} – \int \frac{u}{2} \;du \\ = \frac{u^{2} log\; 2u}{2} – \frac{u^{2}}{4} + C\)
Question 6:
Obtain an integral of u2 log u.
Answer 6:
Suppose, I = \(\int u^{2}\; log\; u \;du\)
Integrating the equation by parts by taking log u as first function and u 2 as second function, we get,
\(I = log\; u \int u ^{2} \; du – \int \left \{ \left ( \frac{d}{du} log\; u \right ) \int u^{2} \;du \right \} \;du \\ = log\; u. \frac{u^{3}}{3} – \int \frac{1}{u}. \frac{u^{2}}{3} \;du \\ = \frac{u^{3} log\; u}{3} – \int \frac{u ^{2}}{3} \;du \\ = \frac{u^{3} log\; u}{2} – \frac{u^{3}}{9} + C\)
Question 7:
Obtain an integral of u sin – 1 u.
Answer 7:
Suppose, I = \(\int u\; sin ^{- 1}\; u \;du\)
Integrating the equation by parts by taking sin – 1 u as first function and u as second function, we get,
\(I = sin ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} sin ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = sin ^{- 1}\; u \frac{u^{2}}{2} – \int \frac{1}{\sqrt{1 – u^{2}}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \frac{- u^{2}}{\sqrt{1 – u^{2}}} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \frac{1 – u^{2}}{\sqrt{1 – u^{2}}} – \frac{1}{\sqrt{1 – u^{2}}} \right \} \;du\) \(= \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \sqrt{1 – u^{2}} – \frac{1}{\sqrt{1 – u^{2}}} \right \} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \left \{ \int \sqrt{1 – u^{2}} \;du – \int \frac{1}{\sqrt{1 – u^{2}}} \;du \right \} \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \left \{ \frac{u}{2} \sqrt{1 – u^{2}} + \frac{1}{2} sin ^{- 1} \;u – sin ^{- 1} \;u \right \} + C \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{u}{4} \sqrt{1 – u^{2}} + \frac{1}{4} sin ^{- 1} \;u – \frac{1}{2} sin ^{- 1} \;u + C \\ = \frac{1}{4} (2u^{2} – 1) sin ^{- 1} \;u + \frac{u}{4} \sqrt{1 – u ^{2}} + C\)
Question 8:
Obtain an integral of u tan – 1 u
Answer 8:
Suppose, I = \(\int u \;tan ^{- 1}\; u\; du\)
Integrating the equation by parts by taking tan – 1 u as first function and u as second function, we get,
\(I = tan ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} tan ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = tan ^{- 1}\; u \;\frac{u^{2}}{2} – \int \frac{1}{1 + u^{2}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \frac{ u^{2}}{1 + u^{2}} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \left \{ \frac{u^{2} + 1}{1 + u^{2}} – \frac{1}{1 + u^{2}} \right \} \;du\) \(= \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \left \{ 1 – \frac{1}{1 + u^{2}} \right \} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int (u – tan ^{- 1} \;u) + C \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{u}{2} + \frac{1}{2}\; tan ^{- 1}\; u + C\)
Question 9:
Obtain an integral of u cos – 1 u
Answer 9:
Suppose, I = \(\int u\; cos ^{- 1}\; u\; du\)
Integrating the equation by parts by taking cos – 1 u as first function and u as second function, we get,
\(I = cos ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} cos ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = cos ^{- 1}\; u \frac{u^{2}}{2} – \int \frac{- 1}{\sqrt{1 – u^{2}}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} – \frac{1}{2} \int \frac{1 – u^{2} + 1}{\sqrt{1 – u^{2}}} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \sqrt{1 – u^{2}} + \frac{- 1}{\sqrt{1 – u^{2}}} \right \} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} – \frac{1}{2} I_{1} – \frac{1}{2}\; cos ^{- 1} \;u ….. (1)\) \(where\; I_{1} = \int \sqrt{1 – u^{2}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{d}{du}\; \sqrt{1 – u^{2}} \int u\; du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{- 2u}{2 \sqrt{1 – u^{2}}}. u \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{- u^{2}}{\sqrt{1 – u^{2}}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{1 – u^{2} – 1}{\sqrt{1 – u^{2}}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \left \{ \int \sqrt{1 – u^{2}} \;du + \int \frac{- du}{\sqrt{1 – u^{2}}} \right \}\) \(I_{1} = u \;\sqrt{1 – u^{2}} – \left \{ I_{1} + cos ^{- 1} \;u \right \} \\ 2 I_{1} = u \;\sqrt{1 – u^{2}} – cos ^{- 1} \;u \\ I_{1} = \frac{u}{2} \;\sqrt{1 – u^{2}} – \frac{1}{2} cos ^{- 1} \;u \\ Using\; in\; equation\; (1),\; we\; get, \\ I = \frac{u^{2} sin ^{- 1}\; u}{2} – \frac{1}{2} \left ( \frac{u}{2}\; \sqrt{1 – u^{2}} – \frac{1}{2}\; cos ^{- 1} \;u \right ) – \frac{1}{2}\; cos ^{- 1} \;u \\ = \frac{(2 u^{2} – 1)}{4}\; cos ^{- 1} \;u – \frac{u}{4}\; \sqrt{1 – u^{2}} + C\)
Question 10:
Obtain an integral of (sin – 1 u) 2
Answer 10:
Suppose, I = \(\int (sin ^{- 1} \;u) ^{2} . 1\; du\)
Integrating the equation by parts by taking (sin – 1 u) 2 as first function and 1 as second function, we get,
\(sin ^{- 1} \;u \int 1 \;du – \int \left \{ \frac{d}{du} (sin ^{- 1} \;u)^{2}. \int 1. du \right \} \;du \\ = (sin ^{- 1} \;u)^{2}. \;u – \int \frac{2\; (sin ^{- 1} \;u)}{\sqrt{1 – u^{2}}}. \;u \;du \\ = u. (sin ^{- 1} \;u)^{2} + \int sin ^{- 1} \;u. \left ( \frac{- 2u}{\sqrt{1 – u^{2}}} \right ) \;du \\ = u. (sin ^{- 1} \;u)^{2} + \left [sin ^{- 1} \;u \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du – \int \left \{ \left ( \frac{d}{du} sin ^{- 1} \;u \right ) \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du\right \} \;du \right ] \\ = u. (sin ^{- 1} \;u)^{2} + \left [ sin ^{- 1} \;u. 2\sqrt{1 – u^{2}} – \int \frac{1}{\sqrt{1 – u^{2}}} . 2 \sqrt{1 – u^{2}} \;du \right ]\) \(= u. (sin ^{- 1} \;u)^{2} + 2 \sqrt{1 – u^{2}}\; sin^{- 1} \;u – \int 2 \;du \\ = u. (sin ^{- 1} \;u)^{2} + 2 \sqrt{1 – u^{2}}\; sin^{- 1} \;u -2u + C\)
Question 11:
Obtain an integral of \(\frac{u\; cos^{- 1} \;u}{\sqrt{1 – u^{2}}}\)
Answer 11:
Suppose, I = \(\int \frac{u\; cos^{- 1} \;u}{\sqrt{1 – u^{2}}} \;du \\ I = \frac{- 1}{2} \int \frac{- 2u}{\sqrt{1 – u^{2}}}.\; cos ^{- 1} u \;du\)
Integrating the equation by parts by taking cos – 1 u as first function and \(\frac{- 2u}{\sqrt{1 – u^{2}}}\) as second function, we get,
\(I = \frac{- 1}{2} \left [cos^{- 1} \;u \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du – \int \left \{ \left ( \frac{d}{du}\; cos^{- 1} \;u \right ) \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du \right \} \;du \right ] \\ = \frac{- 1}{2} \left [ cos^{- 1} \;u. 2\; \sqrt{1 – u^{2}} – \int \frac{- 1}{\sqrt{1 – u^{2}}}. 2\; \sqrt{1 – u^{2}} \;du \right ] \\ = \frac{- 1}{2} \left [2\; \sqrt{1 – u^{2}}\; cos^{- 1} \;u + \int 2 \;du \right ] \\ = \frac{- 1}{2} \left [ 2\; \sqrt{1 – u^{2}}\; cos^{- 1} \;u + 2u \right ] + C \\ = – \left [\sqrt{1 – u^{2}} \;cos^{- 1} \;u + u \right ] + C\)
Question 12:
Obtain an integral of u sec 2 u
Answer 12:
Suppose, I = \(\int u\; sec^{2} \;u \;du\)
Integrating the equation by parts by taking u as first function and sec 2 u as second function, we get,
\(u \int sec^{2} \;u \;du – \int \left \{ \left \{ \frac{d}{du}. u \right \} \int sec^{2} \;u \;du \right \} \;du \\ = u\; tan\; u – \int 1. \;tan \;u \;du \\ = u\; tan\; u – log\; \left | cos\; u \right | + C\)
Question 13:
Obtain an integral of tan – 1 u
Answer 13:
Suppose, I = \(\int tan ^{- 1} \;u \;du\)
Integrating the equation by parts by taking tan – 1 u as first function and 1 as second function, we get,
\(tan ^{- 1} \;u \int 1 \;du – \int \left \{ \left ( \frac{d}{du} \;tan ^{- 1} \;u \right ) \int 1 \;du \right \} \;du \\ = tan ^{- 1} \;u . \;u – \int \frac{1}{1 + u^{2}}. \;u \;du \\ = tan ^{- 1} \;u . \;u – \frac{1}{2} \int \frac{2u}{1 + u^{2}}. \;du \\ = u\; tan ^{- 1} \;u – \frac{1}{2} \;log\; \left | 1 + u^{2} \right | + C \\ = u\; tan ^{- 1} \;u – \frac{1}{2} \;log\; (1 + u^{2}) + C\)
Question 14:
Obtain an integral of u (log u) 2.
Answer 14:
Suppose, I =
Integrating the equation by parts by taking (log u) 2 as first function and 1 as second function, we get,
\(I = (log \;u) ^{2} \int u \;du – \int \left [ \left \{ (\frac{d}{du} \; log \;u) ^{2} \right \} \int u\; du \right ] \;du \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \left [ \int 2 log\; u. \frac{1}{u}. \frac{u^{2}}{2} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \int u\; log\; u \;du\)Integrating the equation again by parts, we get,
\(I = \frac{u^{2}}{2} (log \;u) ^{2} \int u \;du – \left [log \;u \int u \;du – \left \{ (\frac{d}{du} \; log \;u) \int u \;du \right \} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \left [\frac{u^{2}}{2} – log\; u – \int \frac{1}{u}. \frac{u^{2}}{2} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \frac{u^{2}}{2} (log \;u) + \frac{1}{2} \int u \;du \\ = \frac{u^{2}}{2} (log \;u)^{2} – \frac{u^{2}}{2} (log \;u) + \frac{u^{2}}{4} + C\)
Question 15:
Obtain an integral of (u 2 + 1) log u
Answer 15:
Suppose, I = \(\int (u ^{2} + 1) log\; u\; du = \int u ^{2}\; log\; u\; du + \int log\; u\; du \\ Suppose,\; I = I_{1} + I_{2} + …… (1) \\ Where,\; I_{1} = \int u ^{2}\; log\; u\; du \;and\; I_{2} = \int log\; u\; du \\ I_{1} = \int u ^{2}\; log\; u\; du\)
Integrating the equation by parts by taking u as first function and u 2 as second function, we get,
\(I_{1} = (log \;u) – \int u^{2} \;du – \int \left \{ \left ( \frac{d}{du} log \;u \right ) \int u ^{2} \;du \right \} \;du \\ = log \;u. \frac{u ^{3}}{3} – \int \frac{1}{u}. \frac{u ^{3}}{3} \;du \\ = \frac{u ^{3}}{3}\; log \;u – \frac{1}{3} (\int u^{2} \;du) \;du \\ = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + C_{1} …. (2) \\ I_{2} = \int log\; u\; du\)Integrating the equation by parts by taking u as first function and u 2 as second function, we get,
\(I_{2} = (log \;u) – \int 1 \;du – \int \left \{ \left ( \frac{d}{du} log \;u \right ) \int 1 \;du \right \} \\ = log \;u. u – \int \frac{1}{u}. u \;du \\ = u\; log \;u – \int 1 \;du \\ = u\; log \;u – u + C_{2} ….. (3)\)Substituting equations (2) and (3) in equation (1), we get,
\(I = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + C_{1} + u\; log \;u – u + C_{2} \\ = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + u\; log \;u – u + (C_{1} + C_{2}) \\ = \left (\frac{u ^{3}}{3} + u \right ) log \;u – \frac{u ^{3}}{9} – C\)
Question 16:
Obtain an integral of e u (sin u + cos u)
Answer 16:
Suppose, I = \(\int e^{u} (sin\; u + cos\; u) \; du \\ Suppose,\; f (u) = sin\; u \\ f’ (u) = cos\; u \\ I = \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du \\ As\; we\; know, \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e^{u} f (u) + C \\ I = e^{u} \;sin\; u + C\)
Question 17:
Obtain an integral of \(\frac{e^{u}}{(1 + u) ^{2}}\)
Answer 17:
Suppose, I = \(\int \frac{u \;e^{u}}{(1 + u) ^{2}} \;du = \int e^{u} \left \{ \frac{u}{(1 + u) ^{2}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1 + u – 1}{(1 + u) ^{2}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1}{1 + u} – \frac{1}{(1 + u) ^{2}} \right \} \;du \\ Suppose,\; f (u) = \frac{1}{1 + u}, \; \; \; f’ (u) = \frac{- 1}{(1 + u) ^{2}} \\ \int \frac{u\; e^{u}}{(1 + u) ^{2}} \;du = \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du \\ As\; we\; know,\; \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u} \;f (u) + C \\ \int \frac{u\; e^{u}}{(1 + u) ^{2}} \;du = \frac{e^{u}}{1 + u} + C\)
Question 18:
Obtain an integral of \(e^{u} \left ( \frac{1 + sin\; u}{1 + cos\; u} \right )\)
Answer 18:
\(e^{u} \left ( \frac{1 + sin\; u}{1 + cos\; u} \right ) \\ = e^{u} \left ( \frac{sin ^{2} \;\frac{u}{2} + cos ^{2} \;\frac{u}{2} + 2 sin \;\frac{u}{2} \;cos\; \frac{u}{2}}{2\; cos ^{2} \;\frac{u}{2}} \right ) \\ = \frac{e^{u} \left ( sin \;\frac{u}{2} + cos \;\frac{u}{2} \right ) ^{2}}{2 cos ^{2} \;\frac{u}{2}} \\ = \frac{1}{2} e^{u} \left ( \frac{sin \;\frac{u}{2} + cos \;\frac{u}{2}}{cos \;\frac{u}{2}} \right ) ^{2} \\ = \frac{1}{2} e^{u} \left [ tan \;\frac{u}{2} + 1 \right ] ^{2} \\\) \(= \frac{1}{2} e^{u} \left [1 + tan \;\frac{u}{2} \right ] ^{2} \\ = \frac{1}{2} e^{u} \left [ 1 + tan ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ = \frac{1}{2} e^{u} \left [ sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = \left [ \frac{1}{2} sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \; \; \; ….. (1) \\ Suppose,\; tan\; \frac{u}{2} = f (u) \; so \; f’ (u) = \frac{1}{2} sec ^{2} \;\frac{u}{2}\)As we know,
\(\int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C\)Considering equation (1), we get,
\(\int \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = e^{u} tan\; \frac{u}{2} + C\)
Question 19:
Obtain an integral of \(e ^{u} \left ( \frac{1}{u} – \frac{1}{u ^{2}} \right )\)
Answer 19:
Suppose, I = \(\int e ^{u} \left ( \frac{1}{u} – \frac{1}{u ^{2}} \right ) \; du \\ Suppose,\; \frac{1}{u} = f (u) \; \; \; f’ (u) = \frac{- 1}{u^{2}} \\ As\; we\; know,\\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C \\ I = \frac{e ^{u}}{u} + C\)
Question 20:
Obtain an integral of \(\frac{(u – 3) e^{u}}{(u – 1) ^{3}}\)
Answer 20:
\(\int e^{u} \left \{ \frac{(u – 3)}{(u – 1) ^{3}} \right \} \;du = \int e^{u} \left \{ \frac{(u – 3)}{(u – 1 – 2) ^{3}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1}{(u – 1) ^{2} – \frac{2}{(u – 1) ^{3}}} \right \} \;du \\ f (u) = \frac{1}{(u – 1) ^{2}} \; \; \; f’ (u) = \frac{- 2}{(u – 1) ^{3}} \\ As\; we\; know, \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C \\ \int e^{u} \left \{ \frac{(u – 3)}{(u – 1) ^{3}} \right \} \;du = \frac{e ^{u}}{(u – 1) ^{2}} + C\)