Mathematics

# The value of $\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \log { \left( \dfrac { 4+3\sin { x } }{ 4+3\cos { x } } \right) dx } }$ is

$0$

##### SOLUTION
As,
$\begin{array}{l}\int_0^a {f\left( x \right)} dx = \int_0^a {f\left( {x - a} \right)} dx\\So,\\I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \\ = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin \left( {\frac{\pi }{2} - x} \right)}}{{4 + 3\cos \left( {\frac{\pi }{2} - x} \right)}}} \right)} \\ = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \end{array}$
$\begin{array}{l}I + I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} dx + \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} dx\\2I = \int_0^{\frac{\pi }{2}} {\left[ {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) + \log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]dx} \\As,\\\log a + \log b = \log ab\\So,\\2I = \int_0^{\frac{\pi }{2}} {\log \left[ {\left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) \times \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]dx} \\ = \int_0^{\frac{\pi }{2}} {\log \left( 1 \right)dx} \\ = \int_0^{\frac{\pi }{2}} {0dx} \\2I = 0\\I = 0\end{array}$

Thus Option C

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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