Evaluate the Following Integral: ∫ a − a Log ( a − Sin θ a + Sin θ ) D θ

प्रश्न

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

बेरीज

उत्तर

\[\text{Let I} = \int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

Consider

\[f\left( - \theta \right) = \log\left( \frac{a - \sin\left( - \theta \right)}{a + \sin\left( - \theta \right)} \right)\]
\[ = \log\left( \frac{a + \sin\theta}{a - \sin\theta} \right) ............\left[ \sin\left( - x \right) = - \sin x \right]\]
\[ = \log \left( \frac{a - \sin\theta}{a + \sin\theta} \right)^{- 1} \]
\[ = - \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right) ..............\left[ \log a^b = b\log a \right]\]
\[ = - f\left( \theta \right)\]

\[\therefore f\left( - \theta \right) = - f\left( \theta \right)\]

\[ \Rightarrow I = \int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta = 0 .................\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]

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Definite Integrals

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Q 30 Q 29 Q 31

पाठ 19: Definite Integrals - Exercise 20.5 [पृष्ठ ९५]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12

पाठ 19 Definite Integrals
Exercise 20.5 | Q 30 | पृष्ठ ९५

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