# Using the technique of integration by parts evaluate the integral

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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ol3i4c5s4hr
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
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,
$$\displaystyle={\frac{{{x}^{{{4}}}}}{{{4}}}}\times{\ln{{\left({x}\right)}}}-\int{\frac{{{x}^{{{4}}}}}{{{4}}}}\times{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{x}^{{{4}}}}}{{{4}}}}{\ln{{\left({x}\right)}}}-{\frac{{{1}}}{{{4}}}}\int{x}^{{{3}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{x}^{{{4}}}}}{{{4}}}}{\ln{{\left({x}\right)}}}-{\frac{{{1}}}{{{4}}}}{\frac{{{x}^{{{4}}}}}{{{4}}}}$$
$$\displaystyle={\frac{{{x}^{{{4}}}}}{{{4}}}}{\left({\ln{{\left({x}\right)}}}-{\frac{{{1}}}{{{4}}}}\right)}+{C}$$
nick1337

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:
$$\int x^{3}\ln xdx=\ln x(\frac{x^{4}}{4})-\int(\frac{x^{4}}{4})(\frac{1}{x})dx$$
$$=\frac{x^{4}\ln x}{4}-\frac{1}{4}\int x^{3}dx$$
$$=\frac{x\ln x}{4}-\frac{x^{4}}{16}+C$$
$$=\frac{x^{4}(4\ln x-1)}{16}+C$$