-2-2 log 2-sin -2-sin - d - -

Since f( θ)=log ( 2 sin θ/2+sin θ )^ 1 = log ( 2 sin θ/2+sin θ )= f (θ) ∴ f(x) is an odd function of x. Therefore, ∫_ π/2^π/2 log ( 2 sin θ/2+sin θ )d θ =0 ...

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Byju's Answer

Standard XIII

Mathematics

Area between Two Curves

∫-π/2π/2 log ...

Question

$\int_{- \frac{π}{2}}^{\frac{π}{2}} l o g (\frac{2 - s i n θ}{2 + s i n θ}) d θ =$

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Solution

The correct option is A \N
$S i n c e f (- θ) = l o g {(\frac{2 - s i n θ}{2 + s i n θ})}^{- 1} = - l o g (\frac{2 - s i n θ}{2 + s i n θ}) = - f (θ)$
$∴ f (x)$ is an odd function of x.
$T h e r e f o r e, \int_{- \frac{π}{2}}^{\frac{π}{2}} l o g (\frac{2 - s i n θ}{2 + s i n θ}) d θ = 0$

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∫π2−π2 log(2−sin θ2+sin θ)dθ=

The correct option is A \NSince f(−θ)=log(2−sin θ2+sin θ)−1=−log(2−sin θ2+sin θ)=−f(θ) ∴ f(x) is an odd function of x. Therefore, ∫π2−π2 log(2−sin θ2+sin θ)dθ=0

$\int_{- \frac{π}{2}}^{\frac{π}{2}} l o g (\frac{2 - s i n θ}{2 + s i n θ}) d θ =$