First question is ∫(0 to  square root pi)xsin(x^2)dx

Let u=x^2, then du=2xdx. The equation is ∫sin(u)/2 du=1/2(-cos(u))=-1/2cos(x^2). We can sub 0 to sqrt pi to equation, that is 1/2-(-1/2)=1

Using integeration by subsitution is a good way to solve the questiion.

We also can use integration by parts, dv=x v=1/2x^2 . u=sin(x^2) du=xconx. By formula udv=uv-∫vdu. By substitude0 to sqrt pi, we will get 1/2-(-1/2)=1.

Second question is ∫(0 to  square root pi)xcos(x^2)dx

Let u=x^2, then du=2xdx. The equation is ∫cos(u)/2 du=1/2(-sin(u))=-1/2sin(x^2). We can sub 0 to sqrt pi to equation, that is 0-0=1

We also can use integration by parts, dv=x v=1/2x^2 . u=cos(x^2) du=-xsinx. By formula udv=uv-∫vdu. By substitude0 to sqrt pi, we will get 0-0=1.

The third question  ∫(0 to  square root pi)x^3sin(x^2)dx. Let u=x^2, then du=2xdx. The equation is ∫usin(u)/2 du. Using integration by parts, u=u,u’=1,v’=sin(u) v=-cos(u). 1/2(-ucons(u)-(-∫cos(u)du))=1/2(sin(x^2)-x^2cos(x^2)).B y substitude0 to sqrt pi, we will get pi/2-0=pi/2.

The method is used by integration by parts and substitution.