Given the function:
f ( x ) = ∫ 0 x tan ( t − x ) d t − ∫ 0 x f ( t ) tan t d t f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t \, dt f ( x ) = ∫ 0 x tan ( t − x ) d t − ∫ 0 x f ( t ) tan t d t
First, we simplify the first integral. Let u = x − t u = x - t u = x − t d u = − d t du = -dt d u = − d t t = 0 t = 0 t = 0 u = x u = x u = x t = x t = x t = x u = 0 u = 0 u = 0
∫ 0 x tan ( t − x ) d t = ∫ x 0 tan ( − u ) ( − d u ) = ∫ x 0 tan u d u = − ∫ 0 x tan u d u = − [ ln ( sec u ) ] 0 x = − ln ( sec x ) = ln ( cos x ) \int_{0}^{x} \tan(t-x) dt = \int_{x}^{0} \tan(-u) (-du) = \int_{x}^{0} \tan u \, du = -\int_{0}^{x} \tan u \, du = -[\ln(\sec u)]_{0}^{x} = -\ln(\sec x) = \ln(\cos x) ∫ 0 x tan ( t − x ) d t = ∫ x 0 tan ( − u ) ( − d u ) = ∫ x 0 tan u d u = − ∫ 0 x tan u d u = − [ ln ( sec u ) ] 0 x = − ln ( sec x ) = ln ( cos x )
So, the equation becomes:
f ( x ) = ln ( cos x ) − ∫ 0 x f ( t ) tan t d t f(x) = \ln(\cos x) - \int_{0}^{x} f(t) \tan t \, dt f ( x ) = ln ( cos x ) − ∫ 0 x f ( t ) tan t d t
Differentiating both sides with respect to x x x
f ′ ( x ) = − sin x cos x − f ( x ) tan x f'(x) = \dfrac{-\sin x}{\cos x} - f(x) \tan x f ′ ( x ) = cos x − sin x − f ( x ) tan x
f ′ ( x ) + f ( x ) tan x = − tan x f'(x) + f(x) \tan x = -\tan x f ′ ( x ) + f ( x ) tan x = − tan x
This is a linear first-order differential equation. The integrating factor (IF) is:
IF = e ∫ tan x d x = e ln ( sec x ) = sec x \text{IF} = e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x IF = e ∫ t a n x d x = e l n ( s e c x ) = sec x
Multiplying the differential equation by sec x \sec x sec x
f ′ ( x ) sec x + f ( x ) sec x tan x = − sec x tan x f'(x) \sec x + f(x) \sec x \tan x = -\sec x \tan x f ′ ( x ) sec x + f ( x ) sec x tan x = − sec x tan x
d d x ( f ( x ) sec x ) = − sec x tan x \dfrac{d}{dx} (f(x) \sec x) = -\sec x \tan x d x d ( f ( x ) sec x ) = − sec x tan x
Integrating both sides with respect to x x x
f ( x ) sec x = − sec x + C f(x) \sec x = -\sec x + C f ( x ) sec x = − sec x + C
To find C C C f ( x ) = ln ( cos x ) − ∫ 0 x f ( t ) tan t d t \displaystyle f(x) = \ln(\cos x) - \int_{0}^{x} f(t) \tan t \, dt f ( x ) = ln ( cos x ) − ∫ 0 x f ( t ) tan t d t x = 0 x = 0 x = 0 f ( 0 ) = ln ( 1 ) − 0 = 0 f(0) = \ln(1) - 0 = 0 f ( 0 ) = ln ( 1 ) − 0 = 0
0 ⋅ sec 0 = − sec 0 + C ⇒ 0 = − 1 + C ⇒ C = 1 0 \cdot \sec 0 = -\sec 0 + C \Rightarrow 0 = -1 + C \Rightarrow C = 1 0 ⋅ sec 0 = − sec 0 + C ⇒ 0 = − 1 + C ⇒ C = 1
Thus, f ( x ) sec x = 1 − sec x f(x) \sec x = 1 - \sec x f ( x ) sec x = 1 − sec x
f ( x ) = cos x − 1 f(x) = \cos x - 1 f ( x ) = cos x − 1
Now, we find the required derivatives:
f ′ ( x ) = − sin x f'(x) = -\sin x f ′ ( x ) = − sin x
f ′ ′ ( x ) = − cos x f''(x) = -\cos x f ′′ ( x ) = − cos x
Evaluating these at the given points:
f ( π 6 ) = cos ( π 6 ) − 1 = 3 2 − 1 f\left(\dfrac{\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right) - 1 = \dfrac{\sqrt{3}}{2} - 1 f ( 6 π ) = cos ( 6 π ) − 1 = 2 3 − 1
f ′ ( − π 6 ) = − sin ( − π 6 ) = sin ( π 6 ) = 1 2 f'\left(-\dfrac{\pi}{6}\right) = -\sin\left(-\dfrac{\pi}{6}\right) = \sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2} f ′ ( − 6 π ) = − sin ( − 6 π ) = sin ( 6 π ) = 2 1
f ′ ′ ( π 6 ) = − cos ( π 6 ) = − 3 2 f''\left(\dfrac{\pi}{6}\right) = -\cos\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2} f ′′ ( 6 π ) = − cos ( 6 π ) = − 2 3
Finally, substituting these values into the required expression:
f ′ ′ ( π 6 ) + 12 f ′ ( − π 6 ) + f ( π 6 ) = − 3 2 + 12 ( 1 2 ) + ( 3 2 − 1 ) f''\left(\dfrac{\pi}{6}\right) + 12f'\left(-\dfrac{\pi}{6}\right) + f\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2} + 12\left(\dfrac{1}{2}\right) + \left(\dfrac{\sqrt{3}}{2} - 1\right) f ′′ ( 6 π ) + 12 f ′ ( − 6 π ) + f ( 6 π ) = − 2 3 + 12 ( 2 1 ) + ( 2 3 − 1 )
= − 3 2 + 6 + 3 2 − 1 = 5 = -\dfrac{\sqrt{3}}{2} + 6 + \dfrac{\sqrt{3}}{2} - 1 = 5 = − 2 3 + 6 + 2 3 − 1 = 5
Answer: 5 5 5