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Question: If the value of \(I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}x}}{{1 + {a^x}}}dx} \), a>0, the...

If the value of I=∫−ππsin⁡2x1+axdxI = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}x}}{{1 + {a^x}}}dx} , a>0, then II equals?

Explanation

Solution

Hint: If we assume the value of IIto be I=∫abf(x)dx=∫abf(a+b−x)dxI = \int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)} } dx, using the property we will put the value of xx in the equation given in the question as (a+b−x)\left( {a + b - x} \right) which in this case is (−π+π−x)\left( { - \pi + \pi - x} \right)and we get this equation: I=∫−ππsin⁡2(−x)1+a−xdxI = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}\left( { - x} \right)}}{{1 + {a^{ - x}}}}dx} .

Complete step-by-step answer:
Now, in order to solve this question further, we will convert the negative value in the denominator i.e. a−x{a^{ - x}}into positive, making it 1ax\dfrac{1}{{{a^x}}}and replace it in the above equation and sin⁡2(−x){\sin ^2}\left( { - x} \right)can also be written as sin⁡2x{\sin ^2}x.
We have I=∫−ππsin⁡2(−x)1+1axdxI = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}\left( { - x} \right)}}{{1 + \dfrac{1}{{{a^x}}}}}dx} , taking the LCM of the denominator (1+1ax)\left( {1 + \dfrac{1}{{{a^x}}}} \right) we have (ax+1ax)\left( {\dfrac{{{a^x} + 1}}{{{a^x}}}} \right)
Replacing the denominator in the above equation,
I=∫−ππsin⁡2(−x)ax+1axdxI = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}\left( { - x} \right)}}{{\dfrac{{{a^x} + 1}}{{{a^x}}}}}dx} = I=∫−ππaxsin⁡2(x)ax+1dxI = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\sin }^2}\left( x \right)}}{{{a^x} + 1}}dx}
Now, taking the equation I=∫−ππaxsin⁡2(x)ax+1dxI = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\sin }^2}\left( x \right)}}{{{a^x} + 1}}dx} as equation 1
I=∫−ππaxsin⁡2(x)ax+1dxI = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\sin }^2}\left( x \right)}}{{{a^x} + 1}}dx} →\to equation 1
And taking the equation given in the question as equation 2
I=∫−ππsin⁡2x1+axdxI = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}x}}{{1 + {a^x}}}dx} →\to equation 2
Adding equation 1 and 2 we get,
2I=I=∫−ππaxsin⁡2x1+axdx+∫−ππsin⁡2x1+axdx2I = I = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\sin }^2}x}}{{1 + {a^x}}}dx} + \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}x}}{{1 + {a^x}}}} dx
Solving further,
2I=∫−ππsin⁡2x(1+ax1+ax)dx2I = \int\limits_{ - \pi }^\pi {{{\sin }^2}x\left( {\dfrac{{1 + {a^x}}}{{1 + {a^x}}}} \right)} dx
taking the 2 from the left side to the right side of the equation we have,
I=12∫−ππsin⁡2xdxI = \dfrac{1}{2}\int\limits_{ - \pi }^\pi {{{\sin }^2}xdx}
Using cos⁡2x=1−2sin⁡2x  sin⁡2x=1−cos⁡2x2  \cos 2x = 1 - 2{\sin ^2}x \\\ \\\ {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} \\\ we have,
I=14∫−ππ(1−cos⁡2x)dxI = \dfrac{1}{4}\int\limits_{ - \pi }^\pi {\left( {1 - \cos 2x} \right)} dx
Using the property ∫−aaf(x)dx, f(x)=f(−x), 2∫0af(x)dx  \int\limits_{ - a}^a {f\left( x \right)} dx, \\\ f\left( x \right) = f\left( { - x} \right), \\\ 2\int\limits_0^a {f\left( x \right)} dx \\\
I=14×2∫0π(1−cos⁡2x)dxI = \dfrac{1}{4} \times 2\int\limits_0^\pi {\left( {1 - \cos 2x} \right)} dx
I=12[x−sin⁡2x2]0πI = \dfrac{1}{2}\left[ {x - \dfrac{{\sin 2x}}{2}} \right]_0^\pi
Putting the value of xxas π\pi and 0,
I=12[π−sin⁡2π2−0]0πI = \dfrac{1}{2}\left[ {\pi - \dfrac{{\sin 2\pi }}{2} - 0} \right]_0^\pi
I=π2I = \dfrac{\pi }{2}

Note: Using properties such as ∫−aaf(x)dx, f(x)=f(−x), 2∫0af(x)dx  \int\limits_{ - a}^a {f\left( x \right)} dx, \\\ f\left( x \right) = f\left( { - x} \right), \\\ 2\int\limits_0^a {f\left( x \right)} dx \\\ makes the solution easy and solving the questions by using equation 1 and equation 2 is better.