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JEE Mains · Maths · STD 12 - 7.2 definite integral

\(0 \le x \le \frac{\pi }{2}\) માટે \(\int\limits_0^{{{\sin }^2}\,x} {{{\sin }^{ - 1}}\,\left( {\sqrt t } \right)} dt + \int\limits_0^{{{\cos }^2}\,x} {{{\cos }^{ - 1}}\,\left( {\sqrt t } \right)}\, dt\) મેળવો.

  1. A \(\frac{\pi }{4}\)
  2. B \(0\)
  3. C \(1\)
  4. D \(-\frac{\pi }{4}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\pi }{4}\)

Step-by-step Solution

Detailed explanation

Consider \(\int\limits_0^{{{\sin }^2}x} {{{\sin }^{ - 1}}} (\sqrt t )dt + \int\limits_0^{{{\cos }^2}x} {{{\cos }^{ - 1}}} (\sqrt t )dt\) Let \(1=f(x)\) after integrating and putting the limits. \(f(x)=\sin ^{-1} \sqrt{\sin ^{2} x}(2 \sin x \cos x)-0\)…
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