Question

Choose the correct answer.
The value of ${\int }_0^{\frac{\pi }{2}}log\left(\frac{4+3sinx}{4+3cosx}\right)dx$ is
(a) 2 (b) $\frac{3}{4}$ (c) zero (d) -2

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}log\left(\frac{4+3sinx}{4+3cosx}\right)dx\Rightarrow I={\int }_0^{\frac{\pi }{2}}log\left(\frac{4+3sin(\frac{\pi }{2}-x)}{4+3cos(\frac{\pi }{2}-x)}\right)dx[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]\Rightarrow I={\int }_0^{\frac{\pi }{2}}log\left(\frac{4+3cosx}{4+3sinx}\right)dx[\because sin(\frac{\pi }{2}-x)=\text{cos x and cos}(\frac{\pi }{2}-x)=sinx]$
On adding Eqs. (i) and (ii) we get
$2I={\int }_0^{\frac{\pi }{2}}[log\left(\frac{4+3sinx}{4+3cosx}\right)+log\left(\frac{4+3cosx}{4+3sinx}\right)]dx\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}log(\frac{4+3sinx}{4+3cosx}\times \frac{4+3cosx}{4+3sinx})dx[\because logm+logn=logmn]\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}log1dx\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}0dx$
$I=0$
$\text{Hence the correct option is (c)}$