# 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

## Expert Answers 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).

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