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Calculus / Calculus: Early Transcendentals 1 / Chapter 7.5 / Problem 31E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Preliminary work Use a table of integrals to

Chapter 7, Problem 31E

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QUESTION:

Preliminary work Use a table of integrals to evaluate the following indefinite integrals. These integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.

\(\int \frac{\ln x \sin ^{-1}(\ln x)}{x} d x\)

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QUESTION:

Preliminary work Use a table of integrals to evaluate the following indefinite integrals. These integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.

\(\int \frac{\ln x \sin ^{-1}(\ln x)}{x} d x\)

ANSWER:

Problem 31E

Use a table of integrals to evaluate the following indefinite integrals. These integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.
$\int\limits_{\ }^{\ }\frac{ln\left({x}\right)si{n}^{-1}\left({lnx}\right)}{x}dx$
Solution:-
Step 1
First substituting $ln\ x=t$ ,
Differentiating $ln\ x=t$ , we get $\frac{dx}{x}=dt$
Putting these values in the integral, we get

$I=\int\limits_{\ }^{\ }tsi{n}^{-1}\left({t}\right)dt$
Using integration by parts $\int\limits_{\ }^{\ }uv'\ =uv-\int\limits_{\ }^{\ }u'v$
and letting $u=arcsin(t),\ v'=t$ , then
$u'=\frac{1}{\sqrt{1-{t}^{2}}},\ v=\frac{{t}^{2}}{2}$
Now, integral becomes,
$I=\int\limits_{\ }^{\ }tsi{n}^{-1}\left({t}\right)dt$
$=\frac{{t}^{2}arcsin\left({t}\right)}{2}\ -\frac{1}{2}\int\limits_{\ }^{\ }\frac{{t}^{2}}{\sqrt{1-{t}^{2}}}dt$ ( Using integration by parts)
= $\frac{{t}^{2}arcsin\left({t}\right)}{2}-\frac{1}{2}{I}_{1}$ ( ${I}_{1}=$ $\int\limits_{\ }^{\ }\frac{{t}^{2}}{\sqrt{1-{t}^{2}}}dt$ )

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