The correct option is C π/2F(x) = ∫ e^sin^ 1x (1 x/√(1 x^2))dx = e^sin^ 1x (1/√(1 x^2)√(1 x^2) x/√(1 x^2))dx F(x) = e^sin^ 1x√(1 x^2) + c F(0) = 1+c ⇒ c ...

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Integration by Parts

A.B. С.D.

Question

Let $F (x) = \int e^{s i n^{- 1} x} (1 - \frac{x}{\sqrt{1 - x^{2}}}) d x, a n d F (0) = 1, i f F (\frac{1}{2}) = \frac{k {\sqrt{3 e}}^{\frac{x}{6}}}{π}$ , then k =

\frac{π}{4}

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\frac{π}{6}

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\frac{π}{2}

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\frac{π}{3}

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Solution

The correct option is C $\frac{π}{2}$
$F (x) = \int e^{s i n^{- 1} x} (1 - \frac{x}{\sqrt{1 - x^{2}}}) d x = e^{s i n^{- 1} x} (\frac{1}{\sqrt{1 - x^{2}}} \sqrt{1 - x^{2}} - \frac{x}{\sqrt{1 - x^{2}}}) d x F (x) = e^{s i n^{- 1} x} \sqrt{1 - x^{2}} + c F (0) = 1 + c \Rightarrow c = 0 (∵ F (0) = 1) F (\frac{1}{2}) = e^{\frac{π}{6}} . \frac{\sqrt{3}}{2} = \frac{k \sqrt{3}}{π} e^{\frac{π}{6}} ∴ k = \frac{π}{2}$

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Let F(x)=∫esin−1x(1−x√1−x2)dx,andF(0)=1,ifF(12)=k√3ex6π, then k =

The correct option is C π2F(x)=∫esin−1x(1−x√1−x2)dx=esin−1x(1√1−x2√1−x2−x√1−x2)dxF(x)=esin−1x√1−x2+cF(0)=1+c⇒c=0(∵F(0)=1)F(12)=eπ6.√32=k√3πeπ6∴k=π2

Let $F (x) = \int e^{s i n^{- 1} x} (1 - \frac{x}{\sqrt{1 - x^{2}}}) d x, a n d F (0) = 1, i f F (\frac{1}{2}) = \frac{k {\sqrt{3 e}}^{\frac{x}{6}}}{π}$ , then k =