AP EAMCET · Maths · Definite Integration
\(\int_0^{2 \pi} \frac{x \cos (x)}{1+\cos (x)} d x=\)
- A \(\frac{\pi}{6}\)
- B \(\pi^2\)
- C \(\frac{\pi}{4}\)
- D None of the above.
Answer & Solution
Correct Answer
(D) None of the above.
Step-by-step Solution
Detailed explanation
\(I=\int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x\) On applying property \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\), we get \[ I=\int_0^{2 \pi} \frac{(2 \pi-x) \cos x}{1+\cos x} d x \ldots \text { (ii) }\{\because \cos (2 \pi-x)=\cos x\} \] On adding Eqs. (i) and (ii), we get…
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