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Question: The value of \(\int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}x}}{{1 + {a^x}}}dx,a > 0} \) is (a) \...

The value of ∫−ππcos⁡2x1+axdx,a>0\int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}x}}{{1 + {a^x}}}dx,a > 0} is
(a) 2π2\pi
(b) πa\dfrac{\pi }{a}
(c) π2\dfrac{\pi }{2}
(d) aπa\pi

Explanation

Solution

Hint: Here we have to use definite integral properties and trigonometry identities to solve the given integral.

Complete step-by-step answer:
Let I=∫−ππcos⁡2x1+axdxI = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} ...... (1)
Now we know from the properties of definite integral that ∫−aaf(x)dx=∫−aaf(−a+a−x)dx\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_{ - a}^a {f\left( { - a + a - x} \right)} dx
Applying this, we get
I=∫−ππcos⁡2(−π+π−x)1+a−π+π−xdxI = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}\left( { - \pi + \pi - x} \right)}}{{1 + {a^{ - \pi + \pi - x}}}}dx}
I=∫−ππcos⁡2(−x)1+a−xdxI = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}\left( { - x} \right)}}{{1 + {a^{ - x}}}}dx}
As you know cos⁡(−θ)=cos⁡θ\cos \left( { - \theta } \right) = \cos \theta
⇒I=∫−ππcos⁡2(x)1+a−xdx\Rightarrow I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}\left( x \right)}}{{1 + {a^{ - x}}}}dx}
⇒I=∫−ππcos⁡2(x)1+1axdx\Rightarrow I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}\left( x \right)}}{{1 + \dfrac{1}{{{a^x}}}}}dx}
⇒I=∫−ππaxcos⁡2(x)1+axdx\Rightarrow I = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\cos }^2}\left( x \right)}}{{1 + {a^x}}}dx} …… (2)
Now add equation (1) and (2)
⇒2I=∫−ππcos⁡2x1+axdx+∫−ππaxcos⁡2(x)1+axdx\Rightarrow 2I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} + \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\cos }^2}\left( x \right)}}{{1 + {a^x}}}dx}
⇒2I=∫−ππ(1+ax)cos⁡2(x)1+axdx=∫−ππcos⁡2(x)dx\Rightarrow 2I = \int\limits_{ - \pi }^\pi {\dfrac{{\left( {1 + {a^x}} \right){{\cos }^2}\left( x \right)}}{{1 + {a^x}}}dx} = \int\limits_{ - \pi }^\pi {{{\cos }^2}\left( x \right)dx}
Now you know cos⁡2x{\cos ^2}x is written as (1+cos⁡2x2)\left( {\dfrac{{1 + \cos 2x}}{2}} \right)
2I=∫−ππ(1+cos⁡2x2)dx2I = \int\limits_{ - \pi }^\pi {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)dx}
Now for even functions
⇒I=∫−aaf(x)dx=2∫0af(x)dx\Rightarrow I = \int\limits_{ - a}^a {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} }
Now you know that cos is even function so, this integral is written as
2I=2∫0π(1+cos⁡2x2)dx2I = 2\int\limits_0^\pi {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)dx}
I=12∫0π(1+cos⁡2x)dxI = \dfrac{1}{2}\int\limits_0^\pi {\left( {1 + \cos 2x} \right)dx}
Now apply the integral
⇒I=12[x+sin⁡2x2]0π\Rightarrow I = \dfrac{1}{2}\left[ {x + \dfrac{{\sin 2x}}{2}} \right]_0^\pi
⇒I=12[π+sin⁡2π2−0−sin⁡02]\Rightarrow I = \dfrac{1}{2}\left[ {\pi + \dfrac{{\sin 2\pi }}{2} - 0 - \dfrac{{\sin 0}}{2}} \right]
⇒I=12[π+02−0−02]\Rightarrow I = \dfrac{1}{2}\left[ {\pi + \dfrac{0}{2} - 0 - \dfrac{0}{2}} \right]
⇒I=π2\Rightarrow I = \dfrac{\pi }{2}
So option C is correct.

Note: Always remember integral properties. It will help you a lot in solving the integration problems. Also, be careful with signs. Converting higher power trigonometric functions to lower power is always advisable to ease the solving process.