First slide

Methods of integration

Question

The integral $I = \int \frac{\sin^{2} x \cos^{2} x}{{(\sin^{5} x + \cos^{3} x \sin^{2} x + \sin^{3} x \cos^{2} x + \cos^{5} x)}^{2}}$ is equal to

Moderate

A

$\frac{1}{3 (1 - \tan^{2} x)} + C$
B

$\frac{1}{3 (1 + \tan^{3} x)} + C$
C

$\frac{1}{3 (2 + \tan^{3} x)} + C$
D

$- \frac{1}{3 (1 + \tan^{3} x)} + C$

Solution

$\begin{array}{l} (\sin^{5} x + \cos^{3} x {\sin^{2} x + \sin^{3} x \cos^{2} x + \cos^{5} x)}^{2} \\ = \frac{\sin^{2} x \cos^{2} x}{{(\sin^{3} x + \cos^{3} x)}^{2} {(\sin^{2} x + \cos^{2} x)}^{2}} \\ = \frac{\sin^{2} x \cos^{2} x}{{(\sin^{3} x + \cos^{3} x)}^{2}} \\ = \frac{\tan^{2} x \sec^{2} x}{{(1 + \tan^{3} x)}^{2}} \end{array}$
So $\begin{array}{l} I = \int \frac{\tan^{2} x \sec^{2} x}{{(1 + \tan^{3} x)}^{2}} d x \\ = \int \frac{t^{2}}{{(1 + t^{3})}^{2}} d t (t = \tan x) \\ = \frac{1}{3} \int \frac{d u}{u^{2}} (u = 1 + t^{3}) \\ = - \frac{1}{3 u} + C = - \frac{1}{3 (1 + \tan^{3} x)} + C \end{array}$

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