First slide

Methods of integration

Question

$I_{n} = \int \sin^{n} x d x, then 5 I_{5} - 4 I_{3} isequal to$

Moderate

A

$\sin^{4} x \cos x + C$
B

$\cos^{4} x \sin x + C$
C

$- \sin^{4} x \cos x + C$
D

$- \cos^{4} x \sin x + C$

Solution

$I_{5} = \int \sin^{5} x d x$
$\begin{array}{l} I_{5} = \int \sin^{4} x \sin x d x \\ = \sin^{4} x \int \sin x d x - \int [\frac{d}{d x} (\sin^{4} x) \int \sin x d x] d x \\ Using \int f g d x = f \int g d x - \int (f^{1} \int g d x) d x) \\ = - \sin^{4} x (\cos x) - \int 4 \sin^{3} x \cdot \cos x (- \cos x) d x \end{array}$
$\begin{array}{l} = - \sin^{4} x \cos x + 4 \int \sin^{3} x (1 - \sin^{2} x) d x \\ I_{5} = - \sin^{4} x \cos x + 4 \int \sin^{3} x d x - 4 \int \sin^{5} x d x \\ I_{5} = - \sin^{4} x \cos x + 4 I_{3} - 4 I_{5} \\ 5 I_{5} - 4 I_{3} = - \sin^{4} x \cos x + C \end{array}$

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