AP EAMCET · Maths · Indefinite Integration
If \(I_n=\int \frac{\sin n x}{\cos x} d x\), then \(I_n=\)
- A \(\frac{-2}{n-1} \cos (n-1) x-\mathrm{I}_{n-2}\)
- B \(\frac{2}{n-1} \cos (n-1) x+\mathrm{I}_{n-2}\)
- C \(\frac{-2}{n+1} \sin (n+1) x-\mathrm{I}_{n-2}\)
- D \(\frac{-2}{n+1} \cos (n-1) x-\mathrm{I}_{n-2}\)
Answer & Solution
Correct Answer
(A) \(\frac{-2}{n-1} \cos (n-1) x-\mathrm{I}_{n-2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}= & \int \frac{\sin (n x-x+x)}{\cos x} d x \\ = & \int \frac{\sin [(n-1) x+x]}{\cos x} d x \\ & \int \frac{\sin (n-1) x \cos x+\cos (n-1) x \sin x}{\cos } d x\end{aligned}\)…
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