MHT CET · Maths · Definite Integration
\(\int_{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x=\)
- A \(\left(\frac{\pi}{4}\right)\)
- B \(\pi\)
- C \(\left(\frac{\pi}{2}\right)\)
- D \((2 \pi)\)
Answer & Solution
Correct Answer
(D) \((2 \pi)\)
Step-by-step Solution
Detailed explanation
\(\text {Let } \mathrm{I} =\int_{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x \)
\( =\int_{-1}^3\left(\tan ^{-1}\left(\frac{x^2+1}{x}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x \)
\( =\int_{-1}^3 \frac{\pi}{2} \mathrm{~d} x \quad \ldots\left[\because \cot ^{-1}(x)=\tan ^{-1}\left(\frac{1}{x}\right)\right] \)
\( =\frac{\pi}{2}[x]_{-1}^3 \)
\( =\frac{\pi}{2}(4)\)
\( =\int_{-1}^3\left(\tan ^{-1}\left(\frac{x^2+1}{x}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x \)
\( =\int_{-1}^3 \frac{\pi}{2} \mathrm{~d} x \quad \ldots\left[\because \cot ^{-1}(x)=\tan ^{-1}\left(\frac{1}{x}\right)\right] \)
\( =\frac{\pi}{2}[x]_{-1}^3 \)
\( =\frac{\pi}{2}(4)\)
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