`int_sqrt(pi/2)^sqrt(pi) theta^3 cos(theta^2) d theta` First make a substitution and then use integration by parts to evaluate the integral

You need to use the substitution `theta^2= t` , such that:

`theta^2 = t => 2theta d theta= dt => theta d theta= (dt)/2`

Replacing the variable, yields:

`int_(sqrt(pi/2))^(sqrt pi) theta^3(cos(theta^2)d theta = int_(t_1)^(t_2) t*cos t*(dt)/2`

You need to use the integration by parts such that:

`int udv = uv - int vdu`

`u = t => du = dt`

`dv = cos t => v = sin t`

`int t*cos t = t*sin t - int sin t dt`

`int t*cos t = t*sin t + cos t + C`

Replacing back the variable, yields:

`int_(sqrt(pi/2))^(sqrt pi) theta^3(cos(theta^2)d theta = (theta^2*sin (theta^2) + cos (theta^2))|_(sqrt(pi/2))^(sqrt pi)`

Using the fundamental theorem of integration, yields:

`int_(sqrt(pi/2))^(sqrt pi) theta^3(cos(theta^2)d theta = (pi*sin (pi) + cos (pi) - (pi/2)*sin(pi/2) - cos(pi/2))`

`int_(sqrt(pi/2))^(sqrt pi) theta^3(cos(theta^2)d theta = -1 - pi/2`

Hence, evaluating the integral, using substitution, then integration by parts, yields `int_(sqrt(pi/2))^(sqrt pi) theta^3(cos(theta^2)d theta = (1/2)(-1 - pi/2).`