arindameducationusc
  • arindameducationusc
Examples of integration part1
Calculus1
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{Examples~of~integration} \) check the below tutorial by @hartnn its just awesome.... http://openstudy.com/study#/updates/50960518e4b0d0275a3ccfba
arindameducationusc
  • arindameducationusc
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.
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{For~1.} \) example is already in the tutorial. (check it)

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arindameducationusc
  • arindameducationusc
\(\Large\color{red}{2.} \) \[\int\limits_{}^{}\sin (5x+7)= -\frac{ 1 }{ 5 }\cos(5x+7) +C\] C is the arbitrary constant of integration.
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{3.} \) \[\int\limits_{}^{} e ^{xlog _{e}a} dx, ~a>0, ~a \neq1\] \(\Large\color{blue}{Solution(Soln)} \) \[\int\limits_{}^{}e ^{\log _{e}a ^{x}}dx\] and now using property of log we know \[e ^{\log _{e}}=1\] so we get, \[\int\limits_{}^{}a ^{x}dx\] \[=\frac{ a ^{x} }{ \log _{e}a }+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{Q} \) \[\int\limits_{}^{} 10^{x}dx\] \(\Large\color{blue}{Soln} \) \[=\frac{ 10^{x} }{ \log10 }+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{Q.} \) \[\int\limits_{}^{}e ^{x}a ^{x}dx\] \(\Large\color{blue}{Soln} \) \[=\int\limits_{}^{}(ae)^{x}dx\] \[=\frac{ (ae)^{x} }{ \log(ae) } + C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{4.} \) \(\Large\color{red}{Q} \) \[I=\int\limits_{}^{}\frac{ x^{3}+5x^{2}+4x+1 }{ x^{2} } dx\] \(\Large\color{blue}{Soln} \) =\[\int\limits_{}^{}(x+5+\frac{ 4 }{ x }+\frac{ 1 }{ x^{2} } ) dx\] =\[\int\limits_{}^{}xdx+\int\limits_{}^{}5dx+4\int\limits_{}^{}\frac{ 1 }{ x } dx+\int\limits_{}^{}\frac{ 1 }{ x^{2} }dx\] \[=\frac{ x^{2} }{ 2 }+5x+4\log \left| x \right|-\frac{ 1 }{ x }+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{Q} \) \[I=\int\limits_{}^{}(\sqrt{x}+\frac{ 1 }{ \sqrt{x} })^{2}dx\] \(\Large\color{blue}{Soln} \)\ \[=\int\limits_{}^{}(x+\frac{ 1 }{ x }+2)dx\] \[=\int\limits_{}^{} xdx+\int\limits_{}^{}\frac{ 1 }{ x}dx+2\int\limits_{}^{}1dx\] =\[\frac{ x ^{2} }{ 2 }+\log \left| x \right|+2x+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{5} \) \(\Large\color{red}{Q} \) \[I=\int\limits_{}^{}\sqrt{1-\cos2x}dx\] \(\Large\color{blue}{Soln} \) \[I=\sqrt{1-(1-2\sin ^{2}x)}dx\] \[=\sqrt{2\sin^{2}x}dx\] \[=\sqrt{2}\int\limits_{}^{}sinxdx\] \[=-\sqrt{2}cosx+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{6.} \) \(\Large\color{red}{Q} \) \[I=\int\limits_{}^{}\sqrt{1+\cos2x}dx\] \(\Large\color{blue}{Soln} \) \[=\int\limits_{}^{}\sqrt{1+2\cos ^{2}x-1}dx\] \[=\int\limits_{}^{}\sqrt{2\cos ^{2}x}dx\] \[=\sqrt{2}sinx+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{5~and~6} \) \(\Large\color{red}{Q} \) \[I=\sqrt{1+\sin2x}dx\] \(\Large\color{blue}{Soln} \) \[I=\sqrt{\sin ^{2}x+\cos ^{2}x+2sinxcosx}dx\] \[I=\int\limits_{}^{}\sqrt{(sinx+cosx)^{2}}dx\] \[I=\int\limits_{}^{}(sinx+cosx)dx\] \[I=\int\limits_{}^{}sinxdx+\int\limits_{}^{}cosxdx\] \[=-cosx+sinx+C\]
zzr0ck3r
  • zzr0ck3r
Also, they make calculus books full of examples.
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{7.} \) \(\Large\color{red}{Q} \) \[\int\limits_{}^{}\tan2x-\tan(x-\theta)-\tan(x+\theta)~dx\] \(\Large\color{blue}{Soln} \) \[=-1/2\log \cos2x+logcos(x-\theta)+logcos(x+\theta)+C\] \[=1/2\log \sec2x-logsec(x-\theta)-logsec(x+\theta)+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{8.} \) \(\Large\color{red}{Q} \) \[\int\limits_{}^{}cosa+\cot(x-a)sina~dx\] \(\Large\color{blue}{Soln} \) \[=cosa \int\limits1.dx+sina \int\limits \cot(x-a)dx\] \[=xcosa+sina \log \left| \sin(x-a) \right|+C\] \[=xcosa-sinalog \left| cosec(x-a) \right|+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{9.} \) \(\Large\color{red}{Q} \) \[\int\limits \frac{ 1 }{ \sqrt{1+\cos2x} }dx\] \(\Large\color{blue}{Soln} \) \[=\frac{ 1 }{ \sqrt{2\cos ^{2}x} }dx\] \[=\frac{ 1 }{ \sqrt{2} }\int\limits \frac{ 1 }{ \sqrt{\cos ^{2}x} }dx\] \[=\frac{ 1 }{ \sqrt{2} }\int\limits secx dx \] //as 1/cosx=secx and the square and squareroot cancels. \[=\frac{ 1 }{ \sqrt{2} }\log \left| secx+tanx \right|+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{10.} \) \(\Large\color{red}{Q} \) \[\int\limits \frac{ 1 }{ \sqrt{1-cosx} }\] \(\Large\color{blue}{Soln} \) \[=\int\limits \frac{ 1 }{ \sqrt{2\sin ^{2}\frac{ x }{ 2 }} }dx\] \[\frac{ 1 }{ \sqrt{2} }\int\limits cosec \frac{ x }{ 2 }dx\] // 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). \[=\frac{ \sqrt{2} }{ 1 }\log \left| cosec \frac{ x }{ 2 }-\cot \frac{ x }{ 2 } \right|+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{11.} \) \(\Large\color{red}{Q} \) \[\int\limits \tan ^{2}xdx\] \(\Large\color{blue}{Soln} \) \[=\int\limits \sec ^{2}x-1 dx\] \[=tanx-x+C\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{12.} \) \(\Large\color{red}{Q} \) \[\int\limits \cot ^{2}x dx\] \(\Large\color{blue}{Soln} \) \[=\int\limits cosec ^{2}x-1 dx\] \[=-cotx-x+C\]

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