Calculust-4

Calculus Level 4

Let h ( x ) = ( f ∘ g ) ( x ) + K h(x)= (f \circ g)(x) +K , where K K is any constant.

If d d x h ( x ) = − sin ⁡ x cos ⁡ 2 ( cos ⁡ x ) \dfrac{d}{dx} h(x)= \dfrac{-\sin x}{\cos^2{(\cos {x})}} , then compute the value of j ( 0 ) j(0) , where j ( x ) = ∫ g ( x ) f ( x ) f ( t ) g ( t ) d t j(x)=\int^{f(x)}_{g(x)} \dfrac{f(t)}{g(t)} \,dt and, f f and g g are trigonometric functions. If j ( 0 ) j(0) is of the form a − sec ⁡ b a-\sec{b} , then give your answer as a + b a+b .


The answer is 2.

Rudraksh Shukla by Rudraksh Shukla , 23, India

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1 solution

h ( x ) = ∫ − s i n x d x c o s 2 ( c o s x ) h(x) = \int \frac{-sinx dx}{cos^2(cosx)}

Put cosx=t \text{Put cosx=t}

h ( x ) = ∫ d t c o s 2 t = ∫ ( s e c 2 t ) d t = t a n t + K = t a n ( c o s x ) + K h(x) = \int \frac{dt}{cos^2t} = \int (sec^2t) dt= tant+K = tan(cosx) + K

Comparing with the given value of h ( x ) h(x) we get ,

( f ∘ g ) ( x ) = t a n ( c o s x ) ⟹ { f ( x ) = t a n x g ( x ) = c o s x (f\circ g)(x)=tan(cosx)\implies\begin{cases} f(x)=tanx \\ g(x)=cosx\end{cases}

j ( x ) = ∫ c o s x t a n x t a n t c o s t d t = ∫ c o s x t a n x s e c x t a n x d x = [ s e c x ] c o s x t a n x j(x) = \int_{cosx}^{tanx} \frac{tant}{cost}dt = \int_{cosx}^{tanx} secxtanx dx = [secx]_{cosx}^{tanx}

j ( 0 ) = s e c ( t a n 0 ) − s e c ( c o s 0 ) = s e c 0 − s e c 1 = 1 ⏟ a − s e c 1 ⏟ b j(0) = sec(tan0)-sec(cos0)=sec0-sec1=\color{#3D99F6}{\underbrace{1}_{\text{a}}}-sec\color{#D61F06}{\underbrace{1}_{\text{b}}}

a + b = 2 \boxed{a+b=2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Did the same way! +1!

Rudraksh Shukla - 5 years, 1 month ago

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Great ! I am now loving solving your problems

Aditya Narayan Sharma - 5 years, 1 month ago

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Thank u! Will post more soon!

Rudraksh Shukla - 5 years, 1 month ago

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