I -0-2log4-3 sin x -4-3 cos x dx -

Let I=∫^π/2_0 log (4+3 sinx/4+3 cos x )dx...(1) ⇒ I=∫^π/2_0 log [4+3 sin (π/2 x )/4+3 sin (π/2 x ) ]dx ⇒ I=∫^π/2_0 log (4+3 sinx/4+3 cos x )dx...(2) From (1)+ ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

I =∫0π/2log4+...

Question

$I = \int_{0}^{\frac{π}{2}} l o g (\frac{4 + 3 s i n x}{4 + 3 c o s x}) d x$ ?

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Solution

Let $I = \int_{0}^{\frac{π}{2}} l o g (\frac{4 + 3 s i n x}{4 + 3 c o s x}) d x . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} l o g [\frac{4 + 3 s i n (\frac{π}{2} - x)}{4 + 3 s i n (\frac{π}{2} - x)}] d x \Rightarrow I = \int_{0}^{\frac{π}{2}} l o g (\frac{4 + 3 s i n x}{4 + 3 c o s x}) d x . . . (2)$
From (1)+(2), we get
$2 I = \int^{[l o g (\frac{4 + 3 s i n x}{4 + 3 c o s x} + l o g (\frac{4 + 3 c o s x}{4 + 3 s i n x}))]} d x = \int_{0}^{\frac{π}{2}} l o g [(\frac{4 + 3 s i n x}{4 + 3 c o s x}) \times l o g (\frac{4 + 3 c o s x}{4 + 3 s i n x})] d x = \int_{0}^{\frac{π}{2}} (l o g 1) d x = \int_{0}^{\frac{π}{2}} (0) d x = 0$

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Similar questions

Q. Find

I = \int_{0}^{\frac{π}{2}} l o g (\frac{4 + 3 s i n x}{4 + 3 c o s x}) d x

Q. The value of

\int_{0}^{\frac{π}{2}} log (\frac{4 + 3 sin x}{4 + 3 cos x}) d x

The value of is

A. 2

C. 0

Choose the correct answer.
The value of $\int_{0}^{\frac{π}{2}} l o g (\frac{4 + 3 s i n x}{4 + 3 c o s x}) d x$ is
(a) 2 (b) $\frac{3}{4}$ (c) zero (d) -2

\int_{0}^{\frac{π}{2}} l n (\frac{4 + 3 sin x}{4 + 3 cos x}) d x

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I=∫π20log(4+3sinx4+3cosx)dx?

Let I=∫π20log(4+3sinx4+3cosx)dx...(1)⇒I=∫π20log[4+3sin(π2−x)4+3sin(π2−x)]dx⇒I=∫π20log(4+3sinx4+3cosx)dx...(2) From (1)+(2), we get 2I=∫[log(4+3sinx4+3cosx+log(4+3cosx4+3sinx))]dx=∫π20log[(4+3sinx4+3cosx)×log(4+3cosx4+3sinx)]dx=∫π20(log1)dx=∫π20(0)dx=0

$I = \int_{0}^{\frac{π}{2}} l o g (\frac{4 + 3 s i n x}{4 + 3 c o s x}) d x$ ?