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Calculus / Calculus: Early Transcendentals 1 / Chapter 7.3 / Problem 66E

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

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Asymmetric integrands Evaluate the following integrals.

Chapter 4, Problem 66E

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QUESTION:

Asymmetric integrands Evaluate the following integrals. Consider completing the square.

\(\int \frac{d x}{\sqrt{(x-1)(3-x)}}\)

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QUESTION:

Asymmetric integrands Evaluate the following integrals. Consider completing the square.

\(\int \frac{d x}{\sqrt{(x-1)(3-x)}}\)

ANSWER:

Problem 66E

Asymmetric integrands
Evaluate the following integral. Consider completing the square.

$\int\limits_{\ }^{\ }\frac{dx}{\sqrt{\left({x-1}\right)\left({3-x}\right)}}$

Solution:-
Step 1
To evaluate the integral, we make the substitution as follows
$\sqrt{\left({x-1}\right)\left({3-x}\right)}=\left({x-1}\right)t$
Squaring both sides, we get

${\left({\sqrt{\left({x-1}\right)\left({3-x}\right)}}\right)}^{2}=$ ${\left({x-1}\right)}^{2}{t}^{2}$
$\left({x-1}\right)\left({3-x}\right)={\left({x-1}\right)}^{2}{t}^{2}$
$\left({3-x}\right)={\left({x-1}\right)}^{\ }{t}^{2}$
$3-x=x{t}^{2}-{t}^{2}$
$3+{t}^{2}=x\left({1+{t}^{2}}\right)$
$x=\ \frac{3+{t}^{2}}{1+{t}^{2}}$
To get the value of $\sqrt{\left({x-1}\right)\left({3-x}\right)}$ , we know
$\sqrt{\left({x-1}\right)\left({3-x}\right)}=\left({x-1}\right)t$
= $\left({\frac{3+{t}^{2}}{1+{t}^{2}}-1}\right){t}^{\ }$
= $\left({\frac{3+{t}^{2}-1-{t}^{2}}{1+{t}^{2}}}\right){t}^{\ }$
= $\left({\frac{2t}{1+{t}^{2}}}\right){\ }^{\ }$

To evaluate dx, we differentiate $x=\ \frac{3+{t}^{2}}{1+{t}^{2}}$ ,with respect to t.

$x=\ \frac{3+{t}^{2}}{1+{t}^{2}}$
$\frac{dx}{dt}=\frac{2t}{1+{t}^{2}}-\frac{\left({3+{t}^{2}}\right)2t}{{\left({1+{t}^{2}}\right)}^{2}}$
$\frac{dx}{dt}=\frac{2t\left({1+{t}^{2}}\right)}{{\left({1+{t}^{2}}\right)}^{2}}-\frac{\left({3+{t}^{2}}\right)2t}{{\left({1+{t}^{2}}\right)}^{2}}$
$\frac{dx}{dt}=\frac{2t+2{t}^{3}-6t-2{t}^{3}}{{\left({1+{t}^{2}}\right)}^{2}}$
$\frac{dx}{dt}=\frac{-4t}{{\left({1+{t}^{2}}\right)}^{2}}$
$dx=\frac{-4t}{{\left({1+{t}^{2}}\right)}^{2}}dt$

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