MHT CET · Maths · Definite Integration
The value of the integral \(\int_0^1 \sqrt{\frac{1-x}{1+x}} d x\) is
- A \(\left(\frac{\pi}{2}\right)-1\)
- B \(-1\)
- C \(\left(\frac{\pi}{2}\right)+1\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{\pi}{2}\right)-1\)
Step-by-step Solution
Detailed explanation
\(\int_0^1 \sqrt{\frac{1-x}{1+x}} d x=\int_0^1 \frac{1-x}{\sqrt{1-x^2}} d x=\int_0^1 \frac{d x}{\sqrt{1-x^2}}~+\) \(\int_0^1 \frac{-2 x}{2 \sqrt{1-x^2}} d x \)
\( =\left[\sin ^{-1} x\right]_0^1+\left[\sqrt{1-x^2}\right]_0^1 \)
\( =\left\{\sin ^{-1}(1)-\sin ^{-1}(0)\right\}+\left\{\sqrt{1-1^2}-\sqrt{1-0^2}\right\} \)
\( =\frac{\pi}{2}-0+0-1 \)
\( =\frac{\pi}{2}-1\)
\( =\left[\sin ^{-1} x\right]_0^1+\left[\sqrt{1-x^2}\right]_0^1 \)
\( =\left\{\sin ^{-1}(1)-\sin ^{-1}(0)\right\}+\left\{\sqrt{1-1^2}-\sqrt{1-0^2}\right\} \)
\( =\frac{\pi}{2}-0+0-1 \)
\( =\frac{\pi}{2}-1\)
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