Using the technique of integration by parts evaluate the integral

elvishwitchxyp 2021-12-20 Answered
Using the technique of integration by parts evaluate the integral of the product of function \(\displaystyle{\left[{5}+{5}\right]}\)
a) \(\displaystyle{I}=\int{{\sin}^{{-{1}}}{x}}\ {\left.{d}{x}\right.}\)
b) \(\displaystyle{I}=\int{x}^{{{3}}}{\ln{{x}}}\ {\left.{d}{x}\right.}\)

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Expert Answer

ol3i4c5s4hr
Answered 2021-12-21 Author has 1531 answers
Step 1
a) \(\displaystyle{I}=\int{{\sin}^{{-{1}}}{x}}\ {\left.{d}{x}\right.}\) By using integration of by points
\(\displaystyle{I}=\int{1}\times{{\sin}^{{-{1}}}{x}}\ {\left.{d}{x}\right.}={{\sin}^{{-{1}}}{x}}\int{1}{\left.{d}{x}\right.}-\int{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({{\sin}^{{-{1}}}{x}}\right)}\int{1}{\left.{d}{x}\right.}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={{\sin}^{{-{1}}}{x}}\times{x}-\int{\left[{\frac{{{1}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}\times{x}\right]}{\left.{d}{x}\right.}\)
1) \(\displaystyle={x}{{\sin}^{{-{1}}}{x}}-\int{\frac{{{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={I}_{{{1}}}\) (say) let
\(\displaystyle{I}_{{{1}}}=\int{\frac{{{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}\) let \(\displaystyle{t}={1}-{x}^{{{2}}}\)
\(\displaystyle{\left.{d}{t}\right.}=-{2}{x}{\left.{d}{x}\right.}\)
\(\displaystyle=\int{\frac{{{1}}}{{\sqrt{{{t}}}}}}{\frac{{{\left.{d}{t}\right.}}}{{-{2}}}}\)
\(\displaystyle-{\frac{{{\left.{d}{t}\right.}}}{{{2}}}}={x}{\left.{d}{x}\right.}\)
\(\displaystyle=-{\frac{{{1}}}{{{2}}}}\int{t}^{{-{\frac{{{1}}}{{{2}}}}}}{\left.{d}{t}\right.}\)
\(\displaystyle=-{\frac{{{1}}}{{{2}}}}{\left[{\frac{{{t}^{{{\frac{{{1}}}{{{2}}}}}}}}{{{\frac{{{1}}}{{{2}}}}}}}\right]}=-{t}^{{{\frac{{{1}}}{{{2}}}}}}\)
\(\displaystyle{I}_{{{1}}}=-\sqrt{{{1}-{x}^{{{2}}}}}\) put in (1)
\(\displaystyle{I}={x}{{\sin}^{{-{1}}}{x}}-{\left(-\sqrt{{{1}-{x}^{{{2}}}}}\right)}+{c}\)
Step 2
b) \(\displaystyle{I}=\int{x}^{{{3}}}{\ln{{x}}}\ {\left.{d}{x}\right.}\)
By using integration of by parts
\(\displaystyle{I}=\int{x}^{{{3}}}{\ln{{x}}}\ {\left.{d}{x}\right.}={\ln{{x}}}\int{x}^{{{3}}}{\left.{d}{x}\right.}-\int{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\ln{\int}}{x}^{{{3}}}{\left.{d}{x}\right.}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\ln{{x}}}\times{\frac{{{x}^{{{4}}}}}{{{4}}}}-\int{\left[{\frac{{{1}}}{{{x}}}}\times{\frac{{{x}^{{{4}}}}}{{{4}}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{x}^{{{4}}}{\ln{{x}}}}}{{{4}}}}-\int{\frac{{{x}^{{{3}}}}}{{{4}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{I}={\frac{{{x}^{{{4}}}{\ln{{x}}}}}{{{4}}}}-{\frac{{{x}^{{{4}}}}}{{{16}}}}+{c}\)
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twineg4
Answered 2021-12-22 Author has 3655 answers
Step 1
Given: \(\displaystyle\int{x}^{{{3}}}\times{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}\)
Used formula:
\(\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}\)
\(\displaystyle\mu={\ln{{\left({x}\right)}}}\Rightarrow{d}{u}={\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{d}{v}={x}^{{{3}}}\times{\left.{d}{x}\right.}\Rightarrow{v}={\frac{{{x}^{{{4}}}}}{{{4}}}}\)
So,
\(\displaystyle={\frac{{{x}^{{{4}}}}}{{{4}}}}\times{\ln{{\left({x}\right)}}}-\int{\frac{{{x}^{{{4}}}}}{{{4}}}}\times{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{x}^{{{4}}}}}{{{4}}}}{\ln{{\left({x}\right)}}}-{\frac{{{1}}}{{{4}}}}\int{x}^{{{3}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{x}^{{{4}}}}}{{{4}}}}{\ln{{\left({x}\right)}}}-{\frac{{{1}}}{{{4}}}}{\frac{{{x}^{{{4}}}}}{{{4}}}}\)
\(\displaystyle={\frac{{{x}^{{{4}}}}}{{{4}}}}{\left({\ln{{\left({x}\right)}}}-{\frac{{{1}}}{{{4}}}}\right)}+{C}\)
0
nick1337
Answered 2021-12-28 Author has 9672 answers

Step 1
Use integration by parts, which states that:
\(\int udv=uv-\int vdu\)
So, for \(\int x^{3}\ln xdx\), let \(u=\ln x\) and \(dv=x^{3}dx\)
These imply that 
\(du=\frac{1}{x}dx\)
and \(v=\frac{x^{4}}{4}\) (obtain these by differentiating u and integrating dv, respectively).
Plugging these into the integration by parts formula, this yields:
\(\int x^{3}\ln xdx=\ln x(\frac{x^{4}}{4})-\int(\frac{x^{4}}{4})(\frac{1}{x})dx\)
\(=\frac{x^{4}\ln x}{4}-\frac{1}{4}\int x^{3}dx\)
\(=\frac{x\ln x}{4}-\frac{x^{4}}{16}+C\)
\(=\frac{x^{4}(4\ln x-1)}{16}+C\)
 

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