# The $ \int_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} $ equals ?

(A) $ 0 $

(B) $ \dfrac{\pi }{4} $

(C) $ \dfrac{{{\pi ^2}}}{4} $

(D) $ \dfrac{{{\pi ^2}}}{2} $

Last updated date: 19th Mar 2023

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**Hint**: In the given question, we are provided a definite integral to solve. The given problem revolves around the concepts and properties of definite integration. The given question requires us to integrate a function of x with respect to x. Indefinite integration gives us a family of curves. Definite integral gives a numeric value.

**:**

__Complete step-by-step answer__The given question requires us to evaluate a definite integral $ \int_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} $ in variable x consisting of rational trigonometric expression.

So, the function to be integrated is $ \dfrac{{x\sin x}}{{1 + {{\cos }^2}x}} $ , the upper limit of integration is $ \pi $ and the lower limit is $ 0 $ .

Now, we know the property of definite integral according to which \[\int_a^b {f\left( x \right)} = \int_a^b {f\left( {a + b - x} \right)} \]. So, we will use this property so as to evaluate the definite integral.

So, we have, $ I = \int_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} - - - - \left( 1 \right) $

Using the property of definite integral \[\int_a^b {f\left( x \right)} = \int_a^b {f\left( {a + b - x} \right)} \], we get,

\[ \Rightarrow I = \int_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)}}{{1 + {{\cos }^2}\left( {\pi - x} \right)}}dx} \]

Now, we know that $ \sin \left( {\pi - x} \right) = \sin x $ and $ \cos \left( {\pi - x} \right) = - \cos x $ .

Simplifying the expression further, we get,

\[ \Rightarrow I = \int_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\left( { - \cos x} \right)}^2}}}dx} \]

\[ \Rightarrow I = \int_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}dx} - - - - \left( 2 \right)\]

Now, adding the equations marked as $ \left( 1 \right) $ and $ \left( 2 \right) $ , we get,

$ \Rightarrow I + I = \int_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} + \int_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}dx} $

Opening the bracket and simplifying the expression, we get,

$ \Rightarrow 2I = \int_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} + \int_0^\pi {\dfrac{{\pi \sin x}}{{1 + {{\cos }^2}x}}dx - \int_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} } $

Cancelling the like terms with opposite signs, we get,

$ \Rightarrow 2I = \int_0^\pi {\dfrac{{\pi \sin x}}{{1 + {{\cos }^2}x}}dx} $

Now, we get the value of integral as,

$ \Rightarrow I = \dfrac{\pi }{2}\int_0^\pi {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx} $

We can solve the integral using the method of substitution.

We substitute the value of $ \cos x $ as t.

So, we have, $ t = \cos x $

So, we will change the limits of integration according to the new variable.

Then, we know that as x approaches zero, the value of t tends to $ 1 $ . Similarly, as x approaches $ \pi $ , the value of t tends to $ - 1 $ .

Differentiating both sides of $ t = \cos x $ , we get,

$ \Rightarrow dt = - \sin xdx $

Now, substituting the value of $ \cos x $ as t and the value of $ \sin xdx $ as $ - dt $ in the integral, we get,

\[ \Rightarrow I = \dfrac{\pi }{2}\int_1^{ - 1} {\dfrac{{ - dt}}{{1 + {t^2}}}} \]

Now, we know that the integration of $ \dfrac{1}{{1 + {x^2}}} $ with respect to x is $ {\tan ^{ - 1}}x $ . So, putting in the upper and lower limit in the integral, we get,

\[ \Rightarrow I = \dfrac{{ - \pi }}{2}\left[ {{{\tan }^{ - 1}}\left( { - 1} \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right]\]

Now, we know that the value of $ {\tan ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{4} $ and $ {\tan ^{ - 1}}\left( { - 1} \right) = - \dfrac{\pi }{4} $ . So, we get,

\[ \Rightarrow I = \dfrac{{ - \pi }}{2}\left[ { - \dfrac{\pi }{4} - \dfrac{\pi }{4}} \right]\]

Simplifying further, we get,

\[ \Rightarrow I = \dfrac{{ - \pi }}{2}\left[ { - \dfrac{\pi }{2}} \right]\]

\[ \Rightarrow I = \dfrac{{{\pi ^2}}}{4}\]

So, we get the value of the integral $ \int_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} $ as $ \left( {\dfrac{{{\pi ^2}}}{4}} \right) $ .

Hence, option (C) is correct.

**So, the correct answer is “Option C”.**

**Note**: One must have a strong grip over integral calculus to solve such a complex question of definite integration. We also should know about the properties of integration so as to attempt this question. One must take care while doing the calculations in order to be sure of the final answer.

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