If -sin-1 x cos-1 x dx - f-1 x -Ax - xf-1 x - 2 -1 -x2- -2-1 -x2 -2x -c- then

∫sin^ 1 xcos^ 1 xdx = ∫[π2sin^ 1 x sin^ 1 x)^2 ]dx Integrating by parts = π2(xsin^ 1 x + √(1 x^2)) (x(sin^ 1 x)^2 + sin^ 1 x√(1 x^2) x ) +c nbsp; = ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

If ∫sin-1 x c...

Question

If $\int {sin}^{- 1} x {cos}^{- 1} x d x = f^{- 1} (x) [A x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + c,$ then

f (x) = sin x

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f (x) = cos x

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A = \frac{π}{4}

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A = \frac{π}{2}

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Solution

The correct option is D $A = \frac{π}{2}$
$\int {sin}^{- 1} x {cos}^{- 1} x d x = \int [\frac{π}{2} {sin}^{- 1} x - {sin}^{- 1} x)^{2}] d x$

Integrating by parts
$= \frac{π}{2} (x {sin}^{- 1} x + \sqrt{1 - x^{2}}) - (x ({sin}^{- 1} x)^{2} + {sin}^{- 1} x \sqrt{1 - x^{2}} - x) + c$
$= {sin}^{- 1} x [\frac{π}{2} x - x {sin}^{- 1} x - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + c$

$∴ f^{- 1} (x) = {sin}^{- 1} x, A = \frac{π}{2}$
$\Rightarrow f (x) = sin x$

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\int {sin}^{- 1} x {cos}^{- 1} x d x = f^{- 1} (x) [\frac{π}{2} x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + C

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If ∫sin−1xcos−1x dx=f−1(x)[Ax−xf−1(x)−2√1−x2]+π2√1−x2+2x+c, then

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

If $\int {sin}^{- 1} x {cos}^{- 1} x d x = f^{- 1} (x) [A x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + c,$ then