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Question: Solve the following: \[\int\limits_0^{\dfrac{\pi }{2}} {(2\log \sin x - \log \sin 2x)dx} \]...

Solve the following:
∫0π2(2log⁡sin⁡x−log⁡sin⁡2x)dx\int\limits_0^{\dfrac{\pi }{2}} {(2\log \sin x - \log \sin 2x)dx}

Explanation

Solution

If f is continuous on [a,b], then the function g is defined by g(x)=∫axf(t)dtg(x) = \int\limits_a^x {f(t)dt} ,a⩽x⩽ba \leqslant x \leqslant b, is continuous on [a,b] and differentiable on (a,b), and g′(x)=f(x)g'(x) = f(x)
If f is continuous on [a,b], then ∫abf(x)dx=F(b)−F(a)\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 ∫abf(x)dx\int\limits_a^b {f(x)dx} which represent the area bounded by the curve y=f(x)y = f(x), the ordinates x=a, x=b and the x-axis.

Complete step by step answer:
Step 1: let,
I=∫0π2(2log⁡sin⁡x−log⁡sin⁡2x)dxI = \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=∫0π2[2log⁡sin⁡x−log⁡(2sin⁡xcos⁡x)]dxI = \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\log (a.b) = \log a + \log b, using this property of logarithm in our solution,we get
I=∫0π2[2log⁡sin⁡x−log⁡(2sin⁡xcos⁡x)]dxI = \int\limits_0^{\dfrac{\pi }{2}} {[2\log \sin x - \log (2\sin x\cos x)]dx}
I=∫0π2[2log⁡sin⁡x−log⁡2−log⁡sin⁡x−log⁡cos⁡x]dxI = \int\limits_0^{\dfrac{\pi }{2}} {[2\log \sin x - \log 2 - \log \sin x - \log \cos x]dx}
I=∫0π2[log⁡sin⁡x−log⁡2−log⁡cos⁡x]dxI = \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=∫0π2[log⁡sin⁡x−log⁡2−log⁡cos⁡x]dxI = \int\limits_0^{\dfrac{\pi }{2}} {[\log \sin x - \log 2 - \log \cos x]dx}
I=∫0π2log⁡sin⁡xdx−∫0π2log⁡2dx−∫0π2log⁡cos⁡xdxI = \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
I1=∫0π2log⁡cos⁡xdx{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx}
From properties of integration we know that,
∫0af(x)dx=∫0af(a−x)dx\int\limits_0^a {f(x)dx} = \int\limits_0^a {f(a - x)} dx
Using this property, we get
I1=∫0π2log⁡cos⁡xdx{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx}
I1=∫0π2log⁡cos⁡(π2−x)dx{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos (\dfrac{\pi }{2} - x)dx}

I1=∫0π2log⁡sin⁡xdx{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx}
Now, substituting the value of I1 in I we get
I=∫0π2log⁡sin⁡xdx−∫0π2log⁡2dx−∫0π2log⁡cos⁡xdxI = \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=∫0π2log⁡sin⁡xdx−∫0π2log⁡2dx−∫0π2log⁡sin⁡xdxI = \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∫0π2log⁡sin⁡xdx\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} , we get
I=∫0π2log⁡sin⁡xdx−∫0π2log⁡2dx−∫0π2log⁡sin⁡xdxI = \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=−∫0π2log⁡2dxI = - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx}
I=−log⁡2∫0π2dxI = - \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} (Log2 is a constant and therefore taking log2 outside the integration)
I=−log⁡2[x]0π2I = - \log 2\left[ x \right]_0^{\dfrac{\pi }{2}}
I=−log⁡2[π2−0]I = - \log 2[\dfrac{\pi }{2} - 0]
After further simplification, we get
I=π2log⁡12I = \dfrac{\pi }{2}\log \dfrac{1}{2}
Hence, the answer is∫0π2(2log⁡sin⁡x−log⁡sin⁡2x)dx=π2log⁡12\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\sin x to cos⁡x\cos x or vice-versa, always use the quadrant method for giving signs (negative or positive).
Use standard result for solving easily like∫0π2log⁡sin⁡xdx=−π2log⁡2=∫0π2log⁡cos⁡xdx\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx = - \dfrac{\pi }{2}} \log 2 = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx}