KCET · Maths · Area Under Curves
The value of \(\int_{-1 / 2}^{1 / 2} \cos ^{-1} x d x\) is
- A \(\pi\)
- B \(\frac{\pi}{2}\)
- C 1
- D \(\frac{\pi^{2}}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
We have, \(\int_{-1 / 2}^{1 / 2} \cos ^{-1} x d x\)
\(=\int_{-1 / 2}^{1 / 2}\left(\pi / 2-\sin ^{-1} x\right) d x\)
\(=\int_{-1 / 2}^{1 / 2} \pi / 2 d x-\int_{-1 / 2}^{1 / 2} \sin ^{-1} x d x\)
\(=\frac{\pi}{2} \int_{-1 / 2}^{1 / 2} d x-0 \quad\left(\because \sin ^{-1} x\right.\) is odd function \()\)
\(=\frac{\pi}{2}[x]_{-1 / 2}^{1 / 2}=\frac{\pi}{2}\left[\frac{1}{2}-(-1 / 2)\right]=\frac{\pi}{2}(1)=\frac{\pi}{2}\)
\(=\int_{-1 / 2}^{1 / 2}\left(\pi / 2-\sin ^{-1} x\right) d x\)
\(=\int_{-1 / 2}^{1 / 2} \pi / 2 d x-\int_{-1 / 2}^{1 / 2} \sin ^{-1} x d x\)
\(=\frac{\pi}{2} \int_{-1 / 2}^{1 / 2} d x-0 \quad\left(\because \sin ^{-1} x\right.\) is odd function \()\)
\(=\frac{\pi}{2}[x]_{-1 / 2}^{1 / 2}=\frac{\pi}{2}\left[\frac{1}{2}-(-1 / 2)\right]=\frac{\pi}{2}(1)=\frac{\pi}{2}\)
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