If - sin-1 x cos-1 xdx - f-1 -2x - xf-1x - 2 -1-x2- - -2-1 - x2 - 2x - C then fx is equal to

∫ sin^ 1 x cos^ 1 xdx = ∫ [π/2 sin^ 1 x (sin^ 1 x)^2]dx ⇒π/2 (x sin^ 1 x + √(1 x^2)) (x(sin^ 1x)^2 +2 sin^ 1 x √(1 x^2) 2 x) + c(Integration by parts) ⇒ si ...

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Byju's Answer

Standard XIII

Mathematics

Integration by Parts

If ∫ sin-1 x ...

Question

If $\int s i n^{- 1} x c o s^{- 1} x d x = f^{- 1} [\frac{π}{2} x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + C$ then f(x) is equal to

sin 3x

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sin 2x

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sin x

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sin 4x

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Solution

The correct option is C sin x
$\int s i n^{- 1} x c o s^{- 1} x d x = \int [\frac{π}{2} s i n^{- 1} x - (s i n^{- 1} x)^{2}] d x \Rightarrow \frac{π}{2} (x s i n^{- 1} x + \sqrt{1 - x^{2}}) - (x (s i n^{- 1} x)^{2} + 2 s i n^{- 1} x \sqrt{1 - x^{2}} - 2 x) + c$ (Integration by parts)
$\Rightarrow s i n^{- 1} x [\frac{π}{2} x - x s i n^{- 1} x - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + c ∴ f^{- 1} (x) = s i n^{- 1} x, f (x) = s i n x$

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If ∫sin−1xcos−1xdx=f−1[π2x−xf−1(x)−2√1−x2]+π2√1−x2+2x+C then f(x) is equal to

The correct option is C sin x∫sin−1xcos−1xdx=∫[π2sin−1x−(sin−1x)2]dx⇒π2(xsin−1x+√1−x2)−(x(sin−1x)2+2sin−1x√1−x2−2x)+c(Integration by parts) ⇒sin−1x[π2x−x sin−1x−2√1−x2]+π2√1−x2+2x+c∴f−1(x)=sin−1x,f(x)=sinx

If $\int s i n^{- 1} x c o s^{- 1} x d x = f^{- 1} [\frac{π}{2} x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + C$ then f(x) is equal to