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# Solve the following: $\int\limits_0^{\dfrac{\pi }{2}} {(2\log \sin x - \log \sin 2x)dx}$

Last updated date: 18th Jul 2024
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Hint: If f is continuous on [a,b], then the function g is defined by $g(x) = \int\limits_a^x {f(t)dt}$,$a \leqslant x \leqslant b$, is continuous on [a,b] and differentiable on (a,b), and $g'(x) = f(x)$
If f is continuous on [a,b], then $\int\limits_a^b {f(x)dx} = F(b) - F(a)$where F is any alternative of f, that is, a function such that F’=f
A definite integral is denoted by $\int\limits_a^b {f(x)dx}$ which represent the area bounded by the curve $y = f(x)$, the ordinates x=a, x=b and the x-axis.

Step 1: let,
$I = \int\limits_0^{\dfrac{\pi }{2}} {(2\log \sin x - \log \sin 2x)dx}$
Now, from trigonometry properties we know that
Sin2x=2SinxCosx
Substituting the value of sin2x, we get
$I = \int\limits_0^{\dfrac{\pi }{2}} {[2\log \sin x - \log (2\sin x\cos x)]dx}$
Step 2: from logarithm property we know that, $\log (a.b) = \log a + \log b$, using this property of logarithm in our solution,we get
$I = \int\limits_0^{\dfrac{\pi }{2}} {[2\log \sin x - \log (2\sin x\cos x)]dx}$
$I = \int\limits_0^{\dfrac{\pi }{2}} {[2\log \sin x - \log 2 - \log \sin x - \log \cos x]dx}$
$I = \int\limits_0^{\dfrac{\pi }{2}} {[\log \sin x - \log 2 - \log \cos x]dx}$
Step 3: opening the closed bracket so that we can integrate each term separately
$I = \int\limits_0^{\dfrac{\pi }{2}} {[\log \sin x - \log 2 - \log \cos x]dx}$
$I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} }$

Now let
${I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx}$
From properties of integration we know that,
$\int\limits_0^a {f(x)dx} = \int\limits_0^a {f(a - x)} dx$
Using this property, we get
${I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx}$
${I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos (\dfrac{\pi }{2} - x)dx}$

${I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx}$
Now, substituting the value of I1 in I we get
$I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} }$
$I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - } \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx}$
Step 4: cancelling the term$\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx}$, we get
$I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - } \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx}$
$I = - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx}$
$I = - \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx}$ (Log2 is a constant and therefore taking log2 outside the integration)
$I = - \log 2\left[ x \right]_0^{\dfrac{\pi }{2}}$
$I = - \log 2[\dfrac{\pi }{2} - 0]$
After further simplification, we get
$I = \dfrac{\pi }{2}\log \dfrac{1}{2}$
Hence, the answer is$\int\limits_0^{\dfrac{\pi }{2}} {(2\log \sin x - \log \sin 2x)dx} = \dfrac{\pi }{2}\log \dfrac{1}{2}$

Note: While converting $\sin x$ to $\cos x$ or vice-versa, always use the quadrant method for giving signs (negative or positive).
Use standard result for solving easily like$\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx = - \dfrac{\pi }{2}} \log 2 = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx}$