First slide

Methods of integration

Question

The value of $\int \frac{\cos^{3} x + \cos^{5} x}{\sin^{2} x + \sin^{4} x} d x$ is

Moderate

A

$\sin x - 6 \tan^{- 1} (\sin x) + c$
B

$\sin x - 2 {(\sin x)}^{- 1} + c$
C

$\sin x - 2 {(\sin x)}^{- 1} - 6 \tan^{- 1} (\sin x) + c$
D

None of these

Solution

$1 = \int \frac{(\cos^{2} x + \cos^{4} x) \cos x d x}{\sin^{2} x + \sin^{4} x}$

$= \int \frac{1 - t^{2} + {(1 - t^{2})}^{2}}{t^{2} + t^{4}} d t,$ where $\sin x = t$

$= \int \frac{t^{4} - 3 t^{2} + 2}{t^{2} (1 + t^{2})} d t$

$= 2 \int \frac{(t^{4} + t^{2}) - 4 (t^{2} + 1) + 6}{t^{2} (t^{2} + 1)} d t$

$= \int (1 - \frac{4}{t^{2}} + 6 (\frac{1}{t^{2}} - \frac{1}{1 + t^{2}})) d t$

$= t + \frac{- 2}{t} - 6 \tan^{- 1} (t) + c$

$= \sin x - 2 {(\sin x)}^{- 1} - 6 \tan^{- 1} (\sin x) + c$

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