SATHEE: Indefinite Integration 3 Question 14

Indefinite Integration 3 Question 14

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

(a) $g^{'} - \frac{π}{2} = 2 π$

(b) $g^{'} - \frac{π}{2} = - 2 π$

(d) $g^{'} \frac{π}{2} = - 2 π$

Passage Based Questions

Let $f (x) = (1 - x)^{2} \sin^{2} x + x^{2}, \forall x \in R$ and

$g (x) = \int_{1}^{x} \frac{2 (t - 1)}{t + 1} - \ln t f (t) d t \forall x \in (1, \infty)$ .

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Answer:

Correct Answer: 14. $(^{*})$

Solution:

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

$\begin{aligned} g^{'} (x) & = 2 \cos 2 x \sin^{- 1} (\sin 2 x) - \cos x \sin^{- 1} (\sin x) \\ g^{'} \frac{π}{2} & = - 2 \sin^{- 1} (0) = 0 \\ g^{'} - \frac{π}{2} & = - 2 \sin^{- 1} (0) = 0 \end{aligned}$

No option is matching.

Indefinite Integration 3 Question 14

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

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Indefinite Integration 3 Question 14

14. If g(x)=∫sin⁡xsin⁡(2x)sin−1⁡(t)dt, then

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