Question

${\int }_0^{\frac{\pi }{2}}ln\left(\frac{4+3\sin x}{4+3\cos x}\right)dx$.

Answer 1

${\int }_0^{\frac{\pi }{2}}ln\left(\frac{4+3\sin x}{4+3\cos x}\right)dx=I$

Using formula ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$\therefore {\int }_0^{\frac{\pi }{2}}ln\left(\frac{4+3\sin (\frac{\pi }{4}-x)}{4+3\cos (\frac{\pi }{2}-x)}\right)=I$

$={\int }_0^{\frac{\pi }{2}}ln\left(\frac{4+3\cos x}{4+3\sin x}\right)=I$

But ${\int }_0^{\frac{\pi }{2}}ln\left(\frac{4+3\cos x}{4+3\sin x}\right)=-I$

$\therefore -I=I$

$\therefore 2I=0$

$I=0$.