If I_n = int_{0}^{pi} frac{sin(nx)}{sin(x)} d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Consider the difference $I_{n+2} - I_n = \int_{0}^{\pi} \frac{\sin((n+2)x) - \sin(nx)}{\sin(x)} dx$. Using the formula $\sin(A) - \sin(B) = 2 \cos

Vedclass · Accepted Answer

$5\pi$

Vedclass · Answer

(B) Consider the difference $I_{n+2} - I_n = \int_{0}^{\pi} \frac{\sin((n+2)x) - \sin(nx)}{\sin(x)} dx$. Using the formula $\sin(A) - \sin(B) = 2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})$,we get $\sin((n+2)x) - \sin(nx) = 2 \cos((n+1)x) \sin(x)$. Thus,$I_{n+2} - I_n = \int_{0}^{\pi} 2 \cos((n+1)x) dx = [\frac{2 \sin((n+1)x)}{n+1}]_{0}^{\pi} = 0$. This implies $I_n = I_{n+2}$. For $n=1$,$I_1 = \int_{0}^{\pi} 1 dx = \pi$. For $n=2$,$I_2 = \int_{0}^{\pi} \frac{\sin(2x)}{\sin(x)} dx = \int_{0}^{\pi} 2 \cos(x) dx = [2 \sin(x)]_{0}^{\pi} = 0$. Since $I_n = I_{n+2}$,all odd terms $I_1, I_3, \dots, I_9$ are equal to $\pi$,and all even terms $I_2, I_4, \dots, I_{10}$ are equal to $0$. Therefore,$\sum_{n=1}^{10} I_n = (I_1 + I_3 + I_5 + I_7 + I_9) + (I_2 + I_4 + I_6 + I_8 + I_{10}) = 5(\pi) + 5(0) = 5\pi$.

If $I_n = \int_{0}^{\pi} \frac{\sin(nx)}{\sin(x)} dx$,then the value of $\sum_{n=1}^{10} I_n$ is-

Similar Questions

$\int_2^4 \frac{\log x^2}{\log x^2+\log (36-12x+x^2)} dx$ is equal to

The value of $\int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \,dt + \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \,dt$ is

If $\int \limits_0^\pi \frac{5^{\cos x}(1+\cos x \cos 3x+\cos^2 x+\cos^3 x \cos 3x) dx}{1+5^{\cos x}} = \frac{k \pi}{16}$,then $k$ is equal to $...........$.

If $\int_{0}^{1} 4 \cot^{-1}(1-x+x^{2}) dx = a \tan^{-1}(2) - b \log_{e}(5)$, where $a, b \in N$, then $(2a+b)$ is equal to :

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module