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

$\sin x - 2 (\sin x)^{- 1} + 5 \tan^{- 1} (\sin x) + C$
D

none of these

Solution

Since the integrands satisfies.
$R (\sin x, - \cos ∣ x) = - R (\sin x, \cos x)$
Therefore, we substitute $\sin x = t,$ so that $\cos x d x = d t$ .
$\begin{array}{l} \int \frac{\cos^{3} x + \cos^{5} x}{\sin^{2} x + \sin^{4} x} d x \\ = \int \frac{\cos x [(1 - \sin^{2} x) + {(1 - \sin^{2} x)}^{2}]}{\sin^{2} x + \sin^{4} x} d x \\ = \int \frac{(1 - t^{2}) (2 - t^{2})}{t^{2} + t^{4}} d t \\ = \int (1 + \frac{2}{t^{2}} - \frac{6}{1 + t^{2}}) d t \\ = t - \frac{2}{t} - 6 \tan^{- 1} t + C \\ = \sin x - 2 (\sin x)^{- 1} - 6 \tan^{- 1} (\sin x) + C . \end{array}$

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