Integration : More Problems

Problem 2: $\int \frac{\ln(x)}{x} \ dx$

Solution :
Substitute,
$y$ = $\ln(x)$
$\implies dy$ = $\large \frac{dx}{x}$
Thus,
$I$ = $\int \frac{\ln(x)}{x}dx$
$\implies I$ = $\int y dy$
$\implies I$ = $\frac{y^{2}}{2} dy$
$\implies I$ = $\frac{(\ln(x))^{2}}{2} + C$

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Problem 3 : $\int x^{2} \sqrt{1-x^{2}}\ dx$

Solution:
Substitute,
$x = \sin(t)$
$\implies dx = \cos(t) \ dt$
Thus, the integral beomes,
$I$ = $\int \sin^{2}t \sqrt{1-sin^{2}t} \cos t \ dt$
$I$ = $\int \sin^{2}t \cos t \cos t \ dt$
$I$ = $\int \sin^{2}t \cos^{2} t \ dt$

Recall:
$\sin(2\theta) = 2 \ \sin \theta \ \cos \theta$
$\implies \sin \theta \ \cos \theta = \frac{\sin(2\theta)}{2}$

$I$ = $\int (\sin t \cos t)^{2} \ dt$
$I$ = $\int (\frac{\sin(2t)}{2})^{2} \ dt$
$I$ = $\frac{1}{4}\int \sin^{2}(2t) \ dt$

Recall:
$\cos (2 \theta) = \cos^{2} \theta - \sin^{2} \theta$
$\cos (2 \theta) = 1 - 2\sin^{2} \theta$
$\sin^{2} \theta = \frac{1 - \cos (2 \theta)}{2}$

$I$ = $\frac{1}{8}\int (1-\cos(4t)) \ dt$
$I$ = $\frac{1}{8}[t-\frac{\sin(4t)}{4}$
$I$ = $\frac{1}{8}[arcsin(x)-x(1-2x^{2}) \sqrt(1-x^{2}) + c$ [After simplifying $\sin(4t)$ using the above two Recall formulae

Integration : Problem 1

Problem : $\large \int (\sqrt{x^{2} - 4})^{3} dx$

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Solution:

For this problems, we use Euler Substitutions,
Substitute, $\sqrt{x^{2} - 4}$ = $t-x$
$\implies \sqrt{x^{2} - 4}+x$ = $t$ —(1)
$\implies {x^{2} - 4}$ = $(t-x)^{2}$
$\implies {x^{2} - 4}$ = $t^{2}-2tx+ x^{2}$
$\implies {- 4}$ = $t^{2}-2tx$
$\implies {x}$ = $\frac{t^{2}+4}{2t}$ —(2)
$\large \implies \sqrt{x^{2} - 4}$ = $\large \frac{t^{2}+4}{2t}$ —(3)
From (2) and using the quotient rule of differentiation,
$\large dx$ = $\large \frac{t^{2}-4}{2t^{2}}dt$
Plugging these in the given problem we get,
$I$ = $\int (\sqrt{x^{2} - 4})^{3} dx$
$I$ = $\int (\frac{t^{2}+4}{2t})^{3} . \frac{t^{2}-4}{2t^{2}}dt$
$I$ = $\int \frac{(t^{2})^{3}+4^{3}+3.(t^{2})^{2}.4+3.(t^{2}).4^{2}}{8t^{3}} . \frac{t^{2}-4}{2t^{2}}dt$
$I$ = $\int \frac {t^{8}-16t^{6} + 96t^{4}-256t^{2}+256}{16t^{5}}$
$I$ = $\int [\frac{t^{3}}{16} - t + \frac{6}{t} - \frac{16}{t^{3}} + \frac{16}{t^{5}}]$$dt$
$I$ = $\frac{t^{4}}{64} - \frac{t^{2}} {2}+ 6\ln|t| + \frac{8}{t^{2}} -\frac{4}{t^{4}} + C$
Put value of $t$ from equation (1) and get the answer.
Regards.