The value of ∫0sin2⁡x sin−1⁡tdt+∫0cos2⁡x cos−1⁡tdt is

A
$π / 2$
B
1
C
$π / 4$
D
none of these

Solution:

$I = f (x)$ after integration and putting the limits

Now $f^{'} (x) = \sin^{- 1} \sqrt{(\sin^{2} x) \cdot} 2 \sin x \cos x + \cos^{- 1} \sqrt{\cos^{2} x} (- 2 \cos x \sin x) - 0$

$∴ f^{'} (x) = 0 .$ .

Hence $f (x) = c$ (constant)

In order to find c, the constant of integration, we evaluate $f (x) at x = π / 4$

$\begin{array}{l} ∴ I = \int_{0}^{1 / 2} \sin^{- 1} \sqrt{t} d t + \int_{0}^{1 / 2} \cos^{- 1} \sqrt{t} d t \\ = \int_{0}^{1 / 2} (\sin^{- 1} \sqrt{t} + \cos^{- 1} \sqrt{t}) d t \\ = \int_{0}^{1 / 2} \frac{π}{2} d t = \frac{π}{4} = c \\ ∴ f (x) = \frac{π}{4} . \end{array}$

The value of $\int_{0}^{\sin^{2} x} \sin^{- 1} \sqrt{t} d t + \int_{0}^{\cos^{2} x} \cos^{- 1} \sqrt{t} d t$ is

Solution:

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