-0-2log cos x d x-A- -log 2B- -2log 2C- -2 2 log 2D- none

The correct option is B π/2log 2Let I=∫_0^π/2log (cos x)dx=∫_0^π/2log [ cos ( π/2 x ) ]dx= ∫_0^π/2log (sin x)dx ∴ 2I=∫_0^π/2[log (cos x)+log (sin x)]dx=∫_ ...

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Standard XII

Mathematics

Inequalities of Integrals

∫0π/2log cos ...

Question

$\int_{0}^{\frac{π}{2}} l o g (c o s x) d x =$

π l o g 2

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- \frac{π}{2} l o g 2

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- \frac{π}{2} 2 l o g 2

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n o n e

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Solution

The correct option is B $- \frac{π}{2} l o g 2$
$L e t I = \int_{0}^{\frac{π}{2}} l o g (c o s x) d x = \int_{0}^{\frac{π}{2}} l o g [c o s (\frac{π}{2} - x)] d x = \int_{0}^{\frac{π}{2}} l o g (s i n x) d x$
$∴ 2 I = \int_{0}^{\frac{π}{2}} [l o g (c o s x) + l o g (s i n x)] d x = \int_{0}^{\frac{π}{2}} l o g (s i n x c o s x) d x$
$= \int_{0}^{\frac{π}{2}} l o g s i n 2 x d x - \int_{0}^{\frac{π}{2}} l o g 2 d x$
log sin 2x is an odd function, then $\int_{0}^{\frac{π}{2}}$ log sin 2x dx = 0 = 0 - log $2 [x]_{0}^{\frac{π}{2}} = - \frac{π}{2} l o g 2$

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