If g x = ∫ sin x sin 2 x sin - 1 t d t , then

Given, g x = ∫ sin x sin 2 x sin - 1 t d t . . . . . . . . ( i ) D.w.r.to x both side g x = ∫ sin ⁡ x sin ⁡ 2 x sin - 1 ⁡ t d t ⇒ g ' x = 2 sin - 1 ⁡ sin ⁡ 2 x × cos ⁡ 2 x - sin - 1 ⁡ sin ⁡ x cos ⁡ x ⇒ g ' π 2 = 0 a n d g ' - π 2 = 0

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDerivatives (Newton- Leibnitz)Easy2 minPYQ_2017

MathematicsEasymultiple choice

If $g (x) = \int_{\sin x}^{\sin 2 x} \sin^{- 1} t d t,$ then

Options:(select one or more)

Answer:

Solution:

Given, $g (x) = \int_{\sin x}^{\sin 2 x} \sin^{- 1} (t) d t . . . . . . . . (i)$

D.w.r.to x both side $g (x) = \int_{\sin x}^{\sin 2 x} \sin^{- 1} t d t \Rightarrow g^{'} (x) = 2 \sin^{- 1} (\sin 2 x) \times \cos 2 x - \sin^{- 1} (\sin x) \cos x$

$\Rightarrow g^{'} (\frac{π}{2}) = 0 a n d g^{'} (- \frac{π}{2}) = 0$

Stream:JEE_ADVSubject:MathematicsTopic:Definite IntegrationSubtopic:Derivatives (Newton- Leibnitz)

⏱ 2mℹ️ Source: PYQ_2017

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