Question

Evaluate the integral\(\)\(\int\limits_{\sqrt {\frac{\pi }{2}} }^{\sqrt \pi } {{\theta ^3}\cos \left( {{\theta ^2}} \right)d\theta } \)

Short answer

We will use the substitute method and integral by parts method to solve the integral

Let \(x = {\theta ^2}\)

Step-by-step solution

Given Data

Given =\({\rm I} = \int\limits_{\sqrt {\frac{\pi }{2}} }^{\sqrt \pi } {{\theta ^3}\cos \left( {{\theta ^2}} \right)d\theta } \)

\(\begin{aligned}{l}\therefore {\rm I} = \int\limits_{\sqrt {\frac{\pi }{2}} }^{\sqrt \pi } {{\theta ^2}\cos \left( {{\theta ^2}} \right).\frac{1}{2}\left( {2\theta d\theta } \right)} \\{\rm I} = \frac{1}{2}\int\limits_{\frac{\pi }{2}}^\pi {x\cos xdx} \end{aligned}\)

 Step 2: Using Integration by Part Method

\({\rm I} = \mu v - \int {vd\mu } \)

Let,\(\begin{aligned}{l}\mu = x\\d\mu = dx\end{aligned}\) \(\begin{aligned}{l}dv = \cos xdx\\v = \sin x\end{aligned}\)

\(\begin{aligned}{l}{\rm I} = \frac{1}{2}\left( {\left( {x\sin x} \right)_{\frac{\pi }{2}}^\pi - \int\limits_{\frac{\pi }{2}}^\pi {\sin xdx} } \right)\\{\rm I} = \frac{1}{2}\left( {\left( {x\sin x} \right)_{\frac{\pi }{2}}^\pi - \left( {\cos x} \right)_{\frac{\pi }{2}}^\pi } \right)\\{\rm I} = \frac{1}{2}\left( {x\sin x + \cos x} \right)_{\frac{\pi }{2}}^\pi \\{\rm I} = \frac{1}{2}\left( {\pi \sin \pi + \cos \pi } \right) - \frac{1}{2}\left( {\frac{\pi }{2}\sin \frac{\pi }{2} + \cos \frac{\pi }{2}} \right)\\{\rm I} = \frac{1}{2}.\left( { - 1} \right) - \frac{1}{2}\left( {\frac{\pi }{2}} \right)\\\therefore {\rm I} = - \frac{1}{2} - \frac{\pi }{4}\end{aligned}\)

Hence, \(\int\limits_{\sqrt {\frac{\pi }{2}} }^{\sqrt \pi } {{\theta ^3}\cos \left( {{\theta ^2}} \right)d\theta } = - \frac{1}{2} - \frac{\pi }{4}\)