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Question: Evaluate the value of \(\int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx\) . A) \(\dfr...

Evaluate the value of ∫0πxsin⁡x1+cos⁡2xdx\int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx .
A) π24\dfrac{{{\pi ^2}}}{4}
B) π22\dfrac{{{\pi ^2}}}{2}
C) π222\dfrac{{{\pi ^2}}}{{2\sqrt 2 }}
D) π242\dfrac{{{\pi ^2}}}{{4\sqrt 2 }}

Explanation

Solution

Let I=∫0πxsin⁡x1+cos⁡2xdxI = \int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx
Then, applying the property ∫abxdx=∫ab(a+b−x)dx\int\limits_a^b {xdx = \int\limits_a^b {\left( {a + b - x} \right)dx}} .
Now add both the equations before property and after property and solve further.
Finally, find I.

Complete step by step solution:
Let I=∫0πxsin⁡x1+cos⁡2xdxI = \int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx … (1)
Now, applying the property ∫abxdx=∫ab(a+b−x)dx\int\limits_a^b {xdx = \int\limits_a^b {\left( {a + b - x} \right)dx} } in equation (1)
∴I=∫0π(π−x)sin⁡x1+cos⁡2(π−x)dx\therefore I = \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx
=∫0π(π−x)sin⁡x1+cos⁡2xdx= \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}} dx … (2)
On adding equation (1) and equation (2), we get
I+I=∫0πxsin⁡x1+cos⁡2xdx+∫0π(π−x)sin⁡x1+cos⁡2xdx ⇒2I=∫0π(π−x+x)sin⁡x1+cos⁡2xdx ⇒2I=∫0ππsin⁡x1+cos⁡2xdx ⇒2I=π∫0πsin⁡x1+cos⁡2xdx  I + I = \int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx + \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}} dx \\\ \Rightarrow 2I = \int\limits_0^\pi {\dfrac{{\left( {\pi - x + x} \right)\sin x}}{{1 + {{\cos }^2}x}}} dx \\\ \Rightarrow 2I = \int\limits_0^\pi {\dfrac{{\pi \sin x}}{{1 + {{\cos }^2}x}}} dx \\\ \Rightarrow 2I = \pi \int\limits_0^\pi {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}} dx \\\
Let cos⁡x=t\cos x = t.
∴−sin⁡xdx=dt\therefore - \sin xdx = dt
Also, at x=0,t=cos⁡(0)=1x = 0,t = \cos \left( 0 \right) = 1 and at x=π,t=cos⁡(π)=−1x = \pi ,t = \cos \left( \pi \right) = - 1 .
So, 2I=−π∫1−1dt1+t22I = - \pi \int\limits_1^{ - 1} {\dfrac{{dt}}{{1 + {t^2}}}}
Applying the property, ∫−aaxdx=−∫a−axdx\int\limits_{ - a}^a {xdx} = - \int\limits_a^{ - a} {xdx} , we get
2I=π∫−11dt1+t22I = \pi \int\limits_{ - 1}^1 {\dfrac{{dt}}{{1 + {t^2}}}}

∴2I=π[tan⁡−1t]−11 ∴2I=π[tan⁡−1(1)−tan⁡−1(−1)] ∴2I=π[tan⁡−1(1)+tan⁡−1(1)] ∴2I=π×2×tan⁡−1(1) ∴2I2=π×π4 ∴I=π24  \therefore 2I = \pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1 \\\ \therefore 2I = \pi \left[ {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( { - 1} \right)} \right] \\\ \therefore 2I = \pi \left[ {{{\tan }^{ - 1}}\left( 1 \right) + {{\tan }^{ - 1}}\left( 1 \right)} \right] \\\ \therefore 2I = \pi \times 2 \times {\tan ^{ - 1}}\left( 1 \right) \\\ \therefore \dfrac{{2I}}{2} = \pi \times \dfrac{\pi }{4} \\\ \therefore I = \dfrac{{{\pi ^2}}}{4} \\\

Thus, option (A) is correct.

Note:
Some properties of definite integration:
∫abf(x)dx=∫ab(a+b−f(x))dx\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {\left( {a + b - f\left( x \right)} \right)dx} }
∫−aaf(x)dx=−∫a−af(x)dx\int\limits_{ - a}^a {f\left( x \right)dx} = - \int\limits_a^{ - a} {f\left( x \right)dx}
∫abkxdx=k∫abxdx\int\limits_a^b {kxdx = } k\int\limits_a^b {xdx}
∫aaf(x)dx=0\int\limits_a^a {f\left( x \right)dx = } 0
∫abf(x)dx=−∫baf(x)dx\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx}