Question

Evaluate:$\int\limits_0^\pi {\left( {{{\sin }^2}\left( {\dfrac{x}{2}} \right) - {{\cos }^2}\left( {\dfrac{x}{2}} \right)} \right)} dx$

Answer 1

The process of calculating the integrals is known as integration. Definite and indefinite integrals are the two types of integrals. Definite integrals will have limits whereas indefinite integrals don’t have limits. The given sum is a definite integral since we have the limit $0 - \pi$. First, we will try to simplify the given function using trigonometry identities and we will integrate it using the integral formula then, we will apply the limit points.

Formula used: Some of the formulas that we will be using in this problem are:
${x^n}dx = n{x^{n - 1}}$
${\sin ^2}\theta = \dfrac{{1 - \cos (2\theta )}}{2}$
${\cos ^2}\theta = \dfrac{{1 + \cos (2\theta )}}{2}$
$\int {cdu} = c\int {du}$
$\int {cdu = cu}$
$\int {\cos (x)} dx = \sin (x)$
$\int {\cos (2x)dx} = \dfrac{{\sin (2x)}}{2}$

Complete step by step answer:
It is given that$\int\limits_0^\pi {\left( {{{\sin }^2}\left( {\dfrac{x}{2}} \right) - {{\cos }^2}\left( {\dfrac{x}{2}} \right)} \right)} dx$. Let us split this function as $\int A dx - \int B dx = \int\limits_0^\pi {{{\sin }^2}\dfrac{x}{2}dx} - \int\limits_0^\pi {{{\cos }^2}\dfrac{x}{2}dx}$
Let us consider $\int A dx = \int {{{\sin }^2}\left( {\dfrac{x}{2}} \right)} dx$
Let$u = \dfrac{x}{2}$, on differentiating this we get$du = \dfrac{{dx}}{2} \Rightarrow 2du = dx$. On substituting this we get,
$\int A dx = \int {{{\sin }^2}(u)2du}$
Using the formula, $\int {cdu} = c\int {du}$we get
$= 2\int {{{\sin }^2}(u)du}$
We know that${\sin ^2}\theta = \dfrac{{1 - \cos (2\theta )}}{2}$. Using this we get
$= 2\int {\dfrac{{1 - \cos (2u)}}{2}du}$
Again, by using the formula $\int {cdu} = c\int {du}$we get
$= \int {1 - \cos (2u)du}$
Let us split this function into two.
$\int {1 - \cos (2u)du} = \int {1du - \int {\cos (2u)du} }$
Using the formula, $\int {cdu = cu}$we get
$= u - \int {\cos (2u)du}$
Let us substitute$v = 2u$. On differentiating this we get$dv = 2du \Rightarrow du = \dfrac{{dv}}{2}$.
$u - \int {\cos (2u)du} = u - \int {\dfrac{{\cos \left( v \right)}}{2}dv}$
Again, using the formula$\int {cdu} = c\int {du}$ we get
$= u - \dfrac{1}{2}\int {\cos (v)} dv$
Now integrating this by using the formula, $\int {\cos (x)} dx = \sin (x)$we get
$= u - \dfrac{1}{2}\left( {\sin (v)} \right)$
Let’s re-substitute $v = 2u$we get
$= u - \dfrac{{\sin (2u)}}{2}$
Now re-substitute $u = \dfrac{x}{2}$we get
Thus,$\int {Adx} = \dfrac{x}{2} - \dfrac{{\sin (x)}}{2}$
Now let us consider, $\int {Bdx} = \int {{{\cos }^2}\left( {\dfrac{x}{2}} \right)} dx$
Let us substitute$u = \dfrac{x}{2}$. On differentiating this we get$du = \dfrac{{dx}}{2} \Rightarrow 2du = dx$.
$\int {{{\cos }^2}\left( {\dfrac{x}{2}} \right)dx} = \int {{{\cos }^2}(u)2du}$
Using the formula, $\int {cdu} = c\int {du}$we get
$= 2\int {{{\cos }^2}(u)du}$
Now using the formula, ${\cos ^2}\theta = \dfrac{{1 + \cos (2\theta )}}{2}$we get
$= 2\int {\dfrac{{1 + \cos (2u)}}{2}du}$
Again, using the formula$\int {cdu} = c\int {du}$ we get
$= \int {1 + \cos (2u)du}$
Let us split this function into two.
$\int {1 + \cos (2u)du} = \int {1du + \int {\cos (2u)du} }$
Using the formula, $\int {cdu = cu}$we get
$= u + \int {\cos (2u)du}$
On integrating this by using the formula,$\int {\cos (2x)dx} = \dfrac{{\sin (2x)}}{2}$ we get
$= u + \dfrac{{\sin (2u)}}{2}$
Now let us re-substitute $u = \dfrac{x}{2}$we get
Thus, $\int {Bdx} = \dfrac{x}{2} + \dfrac{{\sin (x)}}{2}$
Therefore, $\int {Adx} - \int {Bdx} = \dfrac{x}{2} - \dfrac{{\sin (x)}}{2} - \dfrac{x}{2} - \dfrac{{\sin (x)}}{2}$
On simplifying this we get,
$\int {Adx} - \int {Bdx} = - \sin (x)$
Now let us apply the limits,
$\int\limits_0^\pi {{{\sin }^2}\left( {\dfrac{x}{2}} \right)} - {\cos ^2}\left( {\dfrac{x}{2}} \right)dx = - \mathop {\left[ {\sin (x)} \right]}\nolimits_0^\pi$
$= - \left[ {\sin (\pi ) - \sin (0)} \right]$
We know that $\sin (\pi ) = 0\& \sin (0) = 0$
Thus, we get $\int\limits_0^\pi {{{\sin }^2}\left( {\dfrac{x}{2}} \right)} - {\cos ^2}\left( {\dfrac{x}{2}} \right)dx = 0$

Note: Whenever we use the substitution method in differentiation or integration, we have to re-substitute it so that we will get our solution in the given variables. Otherwise, we will get the answers in temporary variables. While applying the limit we have to first apply the upper limit then the lower limit. The difference between them gives us the value of the integrated function.