Let f be thrice differential function andif $$u=$$−f′′$$($$θ$$)$$$$sin$$⁡θ$$+f$$′$$($$θ$$)$$$$cos$$⁡θ and $$v=f$$′′$$($$θ$$)$$$$cos$$⁡θ$$+f$$′$$($$θ$$)$$$$sin$$⁡θ then $$I=$$∫dudθ$$2+dvd$$θ$$21/2d$$θ is equal to

Question

Let f be thrice differential function andif $$u=$$−f′′$$($$θ$$)$$$$sin$$⁡θ$$+f$$′$$($$θ$$)$$$$cos$$⁡θ and $$v=f$$′′$$($$θ$$)$$$$cos$$⁡θ$$+f$$′$$($$θ$$)$$$$sin$$⁡θ then $$I=$$∫dudθ$$2+dvd$$θ$$21/2d$$θ is equal to

Infinity Learn · Accepted Answer

dudθ$$=$$−f′′$$($$θ$$)$$$$cos$$⁡θ−f′′′$$($$θ$$)$$$$sin$$⁡θ$$+f$$′′$$($$θ$$)$$$$cos$$⁡θ−f′$$($$θ$$)$$$$sin$$⁡θ$$=$$−f′′′$$($$θ$$)+f$$′$$($$θ$$)$$$$sin$$⁡θdvdθ$$=f$$′′′$$($$θ$$)$$$$cos$$⁡θ−f′′$$($$θ$$)$$$$sin$$⁡θ$$+f$$′′$$($$θ$$)=$$$$sin$$⁡θ$$+f$$′′$$($$θ$$)$$$$cos$$⁡θdudθ$$2+dvd$$θ$$2=f$$′′′$$($$θ$$)+f$$′$$($$θ$$)2I=$$∫f′′′$$($$θ$$)+f$$′$$($$θ$$)d$$θ$$=f$$′′$$($$θ$$)+f($$θ$$)+C$$

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

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Let f be thrice differential function andif u=−f′′(θ)sin⁡θ+f′(θ)cos⁡θ and v=f′′(θ)cos⁡θ+f′(θ)sin⁡θ then I=∫dudθ2+dvdθ21/2dθ is equal to

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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