Question

π/2
Prove that ∫ (2 log sinx-log sin 2x)dx= - π/2 log 2
0

Answer 1

A bag contains Rs.396 in terms of 50paise, 2rupee and 5rupee coins in the ratio 4:3:2. Find the number of each type of coins

Answer 2

Given:  

$\bold{\int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x = -\frac{\pi}{2} \log 2 \times}$

Solution:

Let us take equation in L.H.S as $I_1$:

$\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$

$\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}}[2 \log \sin x-\log (2 \sin x \cos x)] d x$

$\bold{[\because \sin 2 \theta=2 \sin \theta \cos \theta]}$

$\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}}[2 \log \sin x-\log 2-\log \sin x-\log \cos x] d x$

$\bold{[\because \log (a b)=\log a+\log b]}$

$\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}}[\log \sin x-\log 2-\log \cos x] d x$

$\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log \sin x d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x-\int_{0}^{\frac{\pi}{2}} \log \cos x d x \rightarrow (1)$

Now let us simplify the solution by considering $\int_{0}^{\frac{\pi}{2}} \log \cos x d x = I_2$

$\mathrm{I}_{2}=\int_{0}^{\frac{\pi}{2}} \log \cos x d x$

$[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x]$

$\mathrm{I}_{2}=\int_{0}^{\frac{\pi}{2}} \log \cos \left(\frac{\pi}{2}-x\right) d x$

$\therefore \mathrm{I}_{2}=\int_{0}^{\frac{\pi}{2}} \log \sin x d x$

On substituting $I_2$ in equation (1), we get,

$\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log \sin x d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x-\int_{0}^{\frac{\pi}{2}} \log \sin x d x$

$\mathrm{I}_{1}=-\int_{0}^{\frac{\pi}{2}} \log 2 d x$

Since $\log 2$ is constant, we get,

$\mathrm{I}_{1}=-\log 2 \int_{0}^{\frac{\pi}{2}} d x$

$\mathrm{I}_{1}=-\log 2[x]_{0}^{\frac{\pi}{2}}$

On substituting the limits, we get,

$\mathrm{I}_{1}=-\log 2\left[\frac{\pi}{2}-0\right]$

$\mathrm{I}_{1}=-\log 2 \times \frac{\pi}{2} = R.H.S$

$\int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x = -\frac{\pi}{2} \log 2 \times$

Hence proved.