Indefinite Integration 3 Question 14
14. If $g(x)=\int _{\sin x}^{\sin (2 x)} \sin ^{-1}(t) d t$, then
(a) $g^{\prime}-\frac{\pi}{2}=2 \pi$
(b) $g^{\prime}-\frac{\pi}{2}=-2 \pi$
(c) $g^{\prime} \frac{\pi}{2}=2 \pi$
(d) $g^{\prime} \frac{\pi}{2}=-2 \pi$
Passage Based Questions
Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}, \forall x \in R$ and
$g(x)=\int _1^{x} \frac{2(t-1)}{t+1}-\ln t \quad f(t) d t \forall x \in(1, \infty)$.
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Answer:
Correct Answer: 14. $\left(^{*}\right)$
Solution:
- $g(x)=\int _{\sin x}^{\sin 2 x} \sin ^{-1}(t) d t$
$$ \begin{aligned} g^{\prime}(x) & =2 \cos 2 x \sin ^{-1}(\sin 2 x)-\cos x \sin ^{-1}(\sin x) \\ g^{\prime} \frac{\pi}{2} & =-2 \sin ^{-1}(0)=0 \\ g^{\prime}-\frac{\pi}{2} & =-2 \sin ^{-1}(0)=0 \end{aligned} $$
No option is matching.
