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JEE Mains · Maths · STD 12 - 7.2 definite integral
For \(0 \le x \le \frac{\pi }{2}\), the value of \(\int\limits_0^{{{\sin }^2}\,x} {{{\sin }^{ - 1}}\,\left( {\sqrt t } \right)} dt + \int\limits_0^{{{\cos }^2}\,x} {{{\cos }^{ - 1}}\,\left( {\sqrt t } \right)}\, dt\) equals
- A \(\frac{\pi }{4}\)
- B \(0\)
- C \(1\)
- D \(-\frac{\pi }{4}\)
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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