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