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Question

Question: \(\int _ { 0 } ^ { 2 \pi } \frac { \sin 2 \theta } { a - b \cos \theta } d \theta =\)...

∫02πsin⁡2θa−bcos⁡θdθ=\int _ { 0 } ^ { 2 \pi } \frac { \sin 2 \theta } { a - b \cos \theta } d \theta =

A

1

B

2

C

π4\frac { \pi } { 4 }

D

0

Answer

0

Explanation

Solution

I=∫02πsin⁡2θa−bcos⁡θdθ=∫02πsin⁡(2π−2θ)a−bcos⁡(2π−θ)dθI = \int _ { 0 } ^ { 2 \pi } \frac { \sin 2 \theta } { a - b \cos \theta } d \theta = \int _ { 0 } ^ { 2 \pi } \frac { \sin ( 2 \pi - 2 \theta ) } { a - b \cos ( 2 \pi - \theta ) } d \theta

⇒ I =−∫02πsin⁡2θa−bcos⁡θdθ= - \int _ { 0 } ^ { 2 \pi } \frac { \sin 2 \theta } { a - b \cos \theta } d \theta

⇒2I=0⇒∫02πsin⁡2θa−bcos⁡θdθ=0\Rightarrow 2 I = 0 \Rightarrow \int _ { 0 } ^ { 2 \pi } \frac { \sin 2 \theta } { a - b \cos \theta } d \theta = 0 .