Question

Evaluate the integral\(\begin{aligned}{l}\int\limits_0^\pi {{x^3}} *{\mathop{\rm Sin}\nolimits} x.dx\\\end{aligned}\)

Short answer

\(\)\({\pi ^3} - 6\pi \)is the value of given integral.

Step-by-step solution

Step 1- To Find the integral

Using By parts method to solve this question that isILATE

\(\int\limits_{.0}^\pi {{x^3}*{\mathop{\rm Sin}\nolimits} x.dx = {x^3}\int {{\mathop{\rm Sin}\nolimits} x.dx - \int {(\frac{d}{{dx}}} } } {x^3}\int {{\mathop{\rm Sin}\nolimits} x).dx} \)

= (-)\({x^3}{\mathop{\rm Cos}\nolimits} x + 3\int {{x^2}{\mathop{\rm Cos}\nolimits} x.dx} \)

=(-)\({x^3}{\mathop{\rm Cos}\nolimits} x + 3({x^2}{\mathop{\rm Sin}\nolimits} x - 2\int {x.{\mathop{\rm Sin}\nolimits} x.dx)} \)

=(-)\({x^3}{\mathop{\rm Cos}\nolimits} x + 3{x^2}{\mathop{\rm Sin}\nolimits} x - 6(x{\mathop{\rm Cos}\nolimits} x + \int {{\mathop{\rm Cos}\nolimits} x.dx)} \)\

=(-) \({x^3}{\mathop{\rm Cos}\nolimits} x + 3{x^2}{\mathop{\rm Sin}\nolimits} x - 6x{\mathop{\rm Cos}\nolimits} x + 6{\mathop{\rm Sin}\nolimits} x + c\)

 Step 2 – To put Upper and lower limit

= \( - ({\pi ^3}){\mathop{\rm Cos}\nolimits} \pi + 3{\pi ^2}{\mathop{\rm Sin}\nolimits} \pi - 6\pi {\mathop{\rm Cos}\nolimits} \pi + 6{\mathop{\rm Sin}\nolimits} x - ( - {0^3}{\mathop{\rm Cos}\nolimits} 0 + 3{(0)^2}{\mathop{\rm Sin}\nolimits} 0 - 6(0){\mathop{\rm Cos}\nolimits} 0 + 6{\mathop{\rm Sin}\nolimits} 0\)

=\(\)\({\pi ^3} - 6\pi \)

Hence,\(\int\limits_0^\pi {{x^3}{\mathop{\rm Sin}\nolimits} x.dx = {\pi ^3} - 6\pi } \)