AP EAMCET · Maths · Indefinite Integration
In \(I_n=\int \frac{\sin n x}{\sin x} d x\) for \(n=1,2,3, \ldots\), then \(I_6=\)
- A \(\frac{3}{5} \sin 3 x+\frac{8}{5} \sin ^5 x-\sin x+c\)
- B \(\frac{2}{5} \sin 5 x-\frac{5}{3} \sin ^3 x-2 \sin x+c\)
- C \(\frac{2}{3} \sin 5 x-\frac{8}{3} \sin ^5 x+4 \sin x+c\)
- D \(\frac{2}{5} \sin 5 x-\frac{8}{5} \sin ^3 x+4 \sin x+c\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{5} \sin 5 x-\frac{8}{5} \sin ^3 x+4 \sin x+c\)
Step-by-step Solution
Detailed explanation
Subtracting Eq. (ii) from Eq. (i), we get \[ \begin{aligned} & I_n-I_{n-2}=\int \frac{\{\sin n x-\sin (n-2) x\}}{\sin x} d x \\ & =\int \frac{2 \cos (n-1) x \sin x}{\sin x} d x=\int 2 \cos (n-1) x d x \\ & =\frac{2 \sin (n-1) x}{(n-1)} \end{aligned} \]…
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