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