First slide

Methods of integration

Question

Let $f$ be thrice differential function and
if $u = - f^{''} (θ) \sin θ + f^{'} (θ) \cos θ$ and $v = f^{''} (θ) \cos θ +$
$f^{'} (θ) \sin θ then I = \int {[{(\frac{d u}{d θ})}^{2} + {(\frac{d v}{d θ})}^{2}]}^{1 / 2} d θ$ is equal to

Moderate

A

$f (θ) - f^{''} (θ) + C$
B

$f (θ) + f^{''} (θ) + C$
C

$f^{'} (θ) + f^{''} (θ) + C$
D

$f^{'} (θ) - f^{''} (θ) + C$

Solution

$\begin{array}{l} \frac{d u}{d θ} = - f^{''} (θ) \cos θ - f^{'''} (θ) \sin θ + f^{''} (θ) \\ \cos θ - f^{'} (θ) \sin θ \\ = - (f^{'''} (θ) + f^{'} (θ)) \sin θ \\ \frac{d v}{d θ} = f^{'''} (θ) \cos θ - f^{''} (θ) \sin θ + f^{''} (θ) \\ = \sin θ + f^{''} (θ) \cos θ \end{array}$
$\begin{array}{l} {(\frac{d u}{d θ})}^{2} + {(\frac{d v}{d θ})}^{2} = {(f^{'''} (θ) + f^{'} (θ))}^{2} \\ I = \int (f^{'''} (θ) + f^{'} (θ)) d θ \\ = f^{''} (θ) + f (θ)) + C \end{array}$

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