Question

Examples of integration part1

Answer 1

\(\Large\color{red}{Examples~of~integration} \)

check the below tutorial by @hartnn
its just awesome....
http://openstudy.com/study#/updates/50960518e4b0d0275a3ccfba

Answer 2

referred with this tutorial I will be giving examples.
All \(\Large\color{blue}{Serial~no.} \) are according to the above tutorial.
just see the above tutorial, check the serial number and you can see the example of the formula.

Answer 3

\(\Large\color{red}{For~1.} \) example is already in the tutorial. (check it)

Answer 4

\(\Large\color{red}{2.} \)
\[\int\limits_{}^{}\sin (5x+7)= -\frac{ 1 }{ 5 }\cos(5x+7) +C\]

C is the arbitrary constant of integration.

Answer 5

\(\Large\color{red}{3.} \)
\[\int\limits_{}^{} e ^{xlog _{e}a} dx, ~a>0, ~a \neq1\]
\(\Large\color{blue}{Solution(Soln)} \)
\[\int\limits_{}^{}e ^{\log _{e}a ^{x}}dx\]
and now using property of log we know
\[e ^{\log _{e}}=1\]
so we get,
\[\int\limits_{}^{}a ^{x}dx\]
\[=\frac{ a ^{x} }{ \log _{e}a }+C\]

Answer 6

\(\Large\color{red}{Q} \) \[\int\limits_{}^{} 10^{x}dx\]
\(\Large\color{blue}{Soln} \)
\[=\frac{ 10^{x} }{ \log10 }+C\]

Answer 7

\(\Large\color{red}{Q.} \)
\[\int\limits_{}^{}e ^{x}a ^{x}dx\]
\(\Large\color{blue}{Soln} \)
\[=\int\limits_{}^{}(ae)^{x}dx\]
\[=\frac{ (ae)^{x} }{ \log(ae) } + C\]

Answer 8

\(\Large\color{red}{4.} \)
\(\Large\color{red}{Q} \)
\[I=\int\limits_{}^{}\frac{ x^{3}+5x^{2}+4x+1 }{ x^{2} } dx\]
\(\Large\color{blue}{Soln} \)
=\[\int\limits_{}^{}(x+5+\frac{ 4 }{ x }+\frac{ 1 }{ x^{2} } ) dx\]
=\[\int\limits_{}^{}xdx+\int\limits_{}^{}5dx+4\int\limits_{}^{}\frac{ 1 }{ x } dx+\int\limits_{}^{}\frac{ 1 }{ x^{2} }dx\]
\[=\frac{ x^{2} }{ 2 }+5x+4\log \left| x \right|-\frac{ 1 }{ x }+C\]

Answer 9

\(\Large\color{red}{Q} \)
\[I=\int\limits_{}^{}(\sqrt{x}+\frac{ 1 }{ \sqrt{x} })^{2}dx\]
\(\Large\color{blue}{Soln} \)\
\[=\int\limits_{}^{}(x+\frac{ 1 }{ x }+2)dx\]
\[=\int\limits_{}^{} xdx+\int\limits_{}^{}\frac{ 1 }{ x}dx+2\int\limits_{}^{}1dx\]
=\[\frac{ x ^{2} }{ 2 }+\log \left| x \right|+2x+C\]

Answer 10

\(\Large\color{red}{5} \)
\(\Large\color{red}{Q} \)
\[I=\int\limits_{}^{}\sqrt{1-\cos2x}dx\]
\(\Large\color{blue}{Soln} \)
\[I=\sqrt{1-(1-2\sin ^{2}x)}dx\]
\[=\sqrt{2\sin^{2}x}dx\]
\[=\sqrt{2}\int\limits_{}^{}sinxdx\]
\[=-\sqrt{2}cosx+C\]

Answer 11

\(\Large\color{red}{6.} \)
\(\Large\color{red}{Q} \)
\[I=\int\limits_{}^{}\sqrt{1+\cos2x}dx\]
\(\Large\color{blue}{Soln} \)
\[=\int\limits_{}^{}\sqrt{1+2\cos ^{2}x-1}dx\]
\[=\int\limits_{}^{}\sqrt{2\cos ^{2}x}dx\]
\[=\sqrt{2}sinx+C\]

Answer 12

\(\Large\color{red}{5~and~6} \)
\(\Large\color{red}{Q} \)
\[I=\sqrt{1+\sin2x}dx\]
\(\Large\color{blue}{Soln} \)
\[I=\sqrt{\sin ^{2}x+\cos ^{2}x+2sinxcosx}dx\]
\[I=\int\limits_{}^{}\sqrt{(sinx+cosx)^{2}}dx\]
\[I=\int\limits_{}^{}(sinx+cosx)dx\]
\[I=\int\limits_{}^{}sinxdx+\int\limits_{}^{}cosxdx\]
\[=-cosx+sinx+C\]

Answer 13

Also, they make calculus books full of examples.

Answer 14

\(\Large\color{red}{7.} \)
\(\Large\color{red}{Q} \)
\[\int\limits_{}^{}\tan2x-\tan(x-\theta)-\tan(x+\theta)~dx\]
\(\Large\color{blue}{Soln} \)
\[=-1/2\log \cos2x+logcos(x-\theta)+logcos(x+\theta)+C\]
\[=1/2\log \sec2x-logsec(x-\theta)-logsec(x+\theta)+C\]

Answer 15

\(\Large\color{red}{8.} \)
\(\Large\color{red}{Q} \)
\[\int\limits_{}^{}cosa+\cot(x-a)sina~dx\]
\(\Large\color{blue}{Soln} \)
\[=cosa \int\limits1.dx+sina \int\limits \cot(x-a)dx\]
\[=xcosa+sina \log \left| \sin(x-a) \right|+C\]
\[=xcosa-sinalog \left| cosec(x-a) \right|+C\]

Answer 16

\(\Large\color{red}{9.} \)
\(\Large\color{red}{Q} \)
\[\int\limits \frac{ 1 }{ \sqrt{1+\cos2x} }dx\]
\(\Large\color{blue}{Soln} \)
\[=\frac{ 1 }{ \sqrt{2\cos ^{2}x} }dx\]
\[=\frac{ 1 }{ \sqrt{2} }\int\limits \frac{ 1 }{ \sqrt{\cos ^{2}x} }dx\]
\[=\frac{ 1 }{ \sqrt{2} }\int\limits secx dx \]
//as 1/cosx=secx and the square and squareroot cancels.
\[=\frac{ 1 }{ \sqrt{2} }\log \left| secx+tanx \right|+C\]

Answer 17

\(\Large\color{red}{10.} \)
\(\Large\color{red}{Q} \)
\[\int\limits \frac{ 1 }{ \sqrt{1-cosx} }\]
\(\Large\color{blue}{Soln} \)
\[=\int\limits \frac{ 1 }{ \sqrt{2\sin ^{2}\frac{ x }{ 2 }} }dx\]
\[\frac{ 1 }{ \sqrt{2} }\int\limits cosec \frac{ x }{ 2 }dx\]
// 1/sin= cosec and squareroot and square cancels.
\[=\frac{ 2 }{ \sqrt{2} }\log \left| cosec \frac{ x }{ 2 }-\cot \frac{ x }{ 2 } \right|+C\]
// denominator of x is 2 so, apply second formula (2).
\[=\frac{ \sqrt{2} }{ 1 }\log \left| cosec \frac{ x }{ 2 }-\cot \frac{ x }{ 2 } \right|+C\]

Answer 18

\(\Large\color{red}{11.} \)
\(\Large\color{red}{Q} \)
\[\int\limits \tan ^{2}xdx\]
\(\Large\color{blue}{Soln} \)
\[=\int\limits \sec ^{2}x-1 dx\]
\[=tanx-x+C\]

Answer 19

\(\Large\color{red}{12.} \)
\(\Large\color{red}{Q} \)
\[\int\limits \cot ^{2}x dx\]
\(\Large\color{blue}{Soln} \)
\[=\int\limits cosec ^{2}x-1 dx\]
\[=-cotx-x+C\]

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