Question

difine integration π limit 0 x sin ²x
dx​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given definite integral.

$\displaystyle \rm = \int^{\pi}_{0}x \sin^{2}(x) \: dx$

We will use integration by parts to solve this problem.

$\rm \longrightarrow u = x$

$\rm \longrightarrow dv= \sin^{2}(x) \: dx$

$\rm \longrightarrow du = dx$

$\rm \longrightarrow v= \dfrac{x}{2} - \dfrac{ \sin(2x) }{4}$

Apply integration by parts:

$\displaystyle \rm \longrightarrow \int u \: dv = uv - \int v \: du$

$\displaystyle \rm =x \bigg( \dfrac{x}{2} -\dfrac{ \sin2x}{4} \bigg) - \int \bigg( \dfrac{x}{2} - \dfrac{ \sin2x}{4} \bigg) \: dx$

$\displaystyle \rm = \dfrac{ {x}^{2} }{2} -\dfrac{x \sin2x}{4} - \int \bigg( \dfrac{x}{2} - \dfrac{ \sin2x}{4} \bigg) \: dx$

$\displaystyle \rm = \dfrac{ {x}^{2} }{2} -\dfrac{x \sin2x}{4} - \int \dfrac{x}{2} \: dx+ \int \dfrac{ \sin2x}{4} \: dx$

$\displaystyle \rm = \dfrac{ {x}^{2} }{2} -\dfrac{x \sin2x}{4} - \dfrac{1}{2} \int x \: dx+ \int \dfrac{ \sin2x}{4} \: dx$

$\displaystyle \rm = \dfrac{ {x}^{2} }{2} -\dfrac{x \sin2x}{4} - \dfrac{ {x}^{2} }{2 \times 2} + \int \dfrac{ \sin2x}{4} \: dx$

$\displaystyle \rm = \dfrac{ {x}^{2} }{2} -\dfrac{x \sin2x}{4} - \dfrac{ {x}^{2} }{4} + \int \dfrac{ \sin2x}{4} \: dx$

$\displaystyle \rm = \dfrac{ {x}^{2} }{2} -\dfrac{x \sin2x}{4} - \dfrac{ {x}^{2} }{4} + \dfrac{1}{4} \int \sin2x \: dx$

$\displaystyle \rm = \dfrac{ {x}^{2} }{2} -\dfrac{x \sin2x}{4} - \dfrac{ {x}^{2} }{4} + \dfrac{1}{4} \times \dfrac{ - \cos2x}{2}$

$\displaystyle \rm = \dfrac{ {x}^{2} }{2} -\dfrac{x \sin2x}{4} - \dfrac{ {x}^{2} }{4} - \dfrac{\cos2x}{8}$

$\displaystyle \rm = \dfrac{ {x}^{2} }{4} -\dfrac{x \sin2x}{4} - \dfrac{\cos2x}{8}$

Now apply the limits:

$\displaystyle \rm = \dfrac{ {x}^{2} }{4} -\dfrac{x \sin2x}{4} - \dfrac{\cos2x}{8} \bigg|^{\pi}_{0}$

$\displaystyle \rm = \bigg( \dfrac{ {\pi}^{2} }{4} -\dfrac{\pi\sin2\pi}{4} - \dfrac{\cos2\pi}{8} \bigg) - \bigg( \dfrac{ {0}^{2} }{4} -\dfrac{0\sin0}{4} - \dfrac{\cos0}{8} \bigg)$

$\displaystyle \rm = \bigg( \dfrac{ {\pi}^{2} }{4}- \dfrac{1}{8} \bigg) - \bigg(- \dfrac{1}{8} \bigg)$

$\displaystyle \rm = \dfrac{ {\pi}^{2} }{4}- \dfrac{1}{8} + \dfrac{1}{8}$

$\displaystyle \rm = \dfrac{ {\pi}^{2} }{4}$

★ Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$