TS EAMCET · Maths · Definite Integration
If \(\left.I_n=\int_{\pi / 2}^{\infty} e^{-x} \cos ^n x d x\right]\), then \(\frac{I_{2018}}{I_{2016}}=\)
- A \(\frac{2018 \times 2019}{(2017)^2+1}\)
- B \(\frac{2018 \times 2017}{(2018)^2+1}\)
- C \(\frac{(2018)(2016)}{(2017)^2+1}\)
- D \(\frac{(2018)(2017)}{(2019)^2+1}\)
Answer & Solution
Correct Answer
(B) \(\frac{2018 \times 2017}{(2018)^2+1}\)
Step-by-step Solution
Detailed explanation
We have, \[ I_n=\int_{\pi / 2}^{\infty} e^{-x} \cos ^n x d x \] Integration by parts…
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