Question

Evaluate: $\int\limits_0^\pi {\dfrac{{4x\sin x}}{{1 + {{\cos }^2}x}}dx} $

Answer 1

Hint: This type of integral is called definite integral with upper and lower limits. The numbers 0 and $\pi $ are called the limits of integration with ‘0’ referred to as the ‘lower limit of integration’ while $\pi $ is referred to as the ‘upper limit of integration’.

Formula used: General formula used to integrate definite integrals is given by
$\int\limits_a^b {f(x)dx = \left[ {F(x)} \right]} _a^b = F(b) - F(a)$
\[\int\limits_a^{ - a} {\dfrac{1}{{1 + {a^2}}}{\text{dt}}} = {\tan ^{ - 1}}\left( a \right)\]

Complete step-by-step answer:
In order to solve this problem we have to equate the question with (variable) \[I\]
\[I = \int\limits_0^\pi {\dfrac{{4x\sin x}}{{1 + {{\cos }^2}x}}dx} {\text{ }} \to {\text{(1)}}\]
Using properties of definite integral, we change the following value of $x$
\[x \to \pi - x\]
Therefore, ‘I’ in the equation becomes,
\[I = 4\int\limits_0^\pi {\dfrac{{(\pi - x)\sin (\pi - x)}}{{1 + {{\cos }^2}x}}dx} \]
\[\sin (\pi - x) = \sin \left( x \right)\]
So we can write it as,
\[I = \int\limits_0^\pi {\dfrac{{4(\pi - x)\sin x}}{{1 + {{\cos }^2}x}}dx} \]
On multiplying the numerator terms we get,
\[I = \int\limits_0^\pi {\dfrac{{\left( {4\pi - 4x} \right)\sin x}}{{1 + {{\cos }^2}x}}dx} \to (2)\]
Now on adding the equations \[\left( 1 \right)\]and\[\left( 2 \right)\], we can write it as, \[I + I = \int\limits_0^\pi {\dfrac{{4x\sin x}}{{1 + {{\cos }^2}x}}dx} {\text{ + }}\int\limits_0^\pi {\dfrac{{\left( {4\pi - 4x} \right)\sin x}}{{1 + {{\cos }^2}x}}dx} \]
Now we cancel the same term in the numerator term, while do the addition so we can write it as,
\[2I = 4\pi \int\limits_0^\pi {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx} ,{\text{or}}\]
On dividing $2$on both side we get,
\[I = 2\pi \int\limits_0^\pi {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx} \to (3)\]
Let’s consider, \[\cos x = t\] and differentiating on both sides we get,
\[\sin xdx = - dt\].
When the limit values has also changed,
\[x = 0{\text{ }} \Rightarrow t = cos{\text{ }}0{\text{ }} = 1\]
\[x = \pi {\text{ }} \Rightarrow t = cos{\text{ }}\pi {\text{ }} = - 1\]
Substituting these values in equation \[\left( 3 \right)\], we have
\[\therefore I = 2\pi \int\limits_1^{ - 1} {\dfrac{{ - dt}}{{1 + {t^2}}}} \]
Here we use this formula, \[\int\limits_a^{ - a} {\dfrac{1}{{1 + {a^2}}}{\text{dt}}} = {\tan ^{ - 1}}\left( a \right)\]
We can write it as,
\[I = 2\pi [{\tan ^{ - 1}}{\text{t}}]_{ - 1}^1\]
Substitute the limit value in \['t'\]we get, \[I = 2\pi [{\tan ^{ - 1}}\left( 1 \right) - {\tan ^{ - 1}}\left( { - 1} \right)]\]
Here ${\tan ^{ - 1}}1 = \dfrac{\pi }{4}$
\[I = 2\pi \left[ {\dfrac{\pi }{4} - \left( { - \dfrac{\pi }{4}} \right)} \right]\]
On adding the bracket terms we get, \[I = 2\pi \left( {\dfrac{{2\pi }}{4}} \right)\]
Let us multiply the terms we get, \[I = \dfrac{{4{\pi ^2}}}{4}\]
Now cancel the term we get,

\[I = {\pi ^2}\]

Note: Note that the symbol $\int {} $ simply means indefinite integral. It is the same symbol used previously for the indefinite integral of a function. An integral in this form without upper and lower limits, also called an ‘antiderivative’. It is also interesting to note that the first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals.
Definite integration is an important section of integral calculus from where a good number of questions on this come in various competitive exams.
We must be aware of the various properties of definite integral in order to solve the problem.
\[1.\int_a^b {f(x)dx = \int_b^a {f(x)dx} } \]
\[2.\int_a^a {f(x)dx = 0} \]
\[3.\int_a^b {cdx} = c(b - a)\], where ‘c’ is any constant
\[4.\int_a^b {cf(x)dx} = c\int_a^b {f(x)} dx,\], where ‘c’ is any constant
\[{\text{5}}{\text{. }}\int_a^b {\left[ {f(x) + g(x)} \right]} dx = \int_a^b {f(x)} dx + \int_a^b {g(x)} dx\]
\[{\text{6}}{\text{. }}\int_a^b {\left[ {f(x) - g(x)} \right]} dx = \int_a^b {f(x)} dx - \int_a^b {g(x)} dx\]
\[7.\int_a^b {f(x)} dx + \int_b^c {f(x)} dx = \int_a^c {f(x)} dx\]