MHT CET · Maths · Definite Integration
\(\int_0^1 \sqrt{\frac{1-x}{1+x}} d x=\)
- A \(\frac{\pi}{4}+1\)
- B \(\frac{\pi}{2}+1\)
- C \(\frac{\pi}{4}-1\)
- D \(\frac{\pi}{2}-1\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{2}-1\)
Step-by-step Solution
Detailed explanation
\(\int_0^1 \sqrt{\frac{1-x}{1+x}} d x=\int_0^1 \frac{1-x}{\sqrt{1-x^2}} d x=\int_0^1 \frac{1}{\sqrt{1-x^2}} d x+\) \(\frac{1}{2} \int_0^1 \frac{-2 x}{\sqrt{1-x^2}} d x \)
\( =\left[\sin ^{-1} x\right]_0^1+\left[\sqrt{1-x^2}\right]_0^1 \)
\( =\sin ^{-1}(1)-\sin ^{-1}(0)+\sqrt{1-1^2}-\sqrt{1-0^2} \)
\( =\frac{\pi}{2}-0+0-1 \)
\( =\frac{\pi}{2}-1\)
\( =\left[\sin ^{-1} x\right]_0^1+\left[\sqrt{1-x^2}\right]_0^1 \)
\( =\sin ^{-1}(1)-\sin ^{-1}(0)+\sqrt{1-1^2}-\sqrt{1-0^2} \)
\( =\frac{\pi}{2}-0+0-1 \)
\( =\frac{\pi}{2}-1\)
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