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   $$\int_a^b f(x) dx = \int_\alpha^\beta f(\phi(t))\phi'(t) dt$$
   
   

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   $$\int_a^b fg' dx = [fg]_a^b - \int_a^b f'g dx$$
   
   

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  $$\int_0^{\pi/2} 3\sin^2 x\cos x dx = \int_0^{\pi/2} 3u^2 du = \left[u^3\right]_0^{\pi/2}$$
  
  

  
  

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  $$\int_0^{\pi/2} x\cos x dx = \int_0^{\pi/2} x (\sin x)' dx = x\sin x - \int_0^{\pi/2} (x)' \sin x dx$$
  
  

  
  

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$$\int_a^b f(x) dx = \left[\int f(x) dx\right]_{x=a}^{x=b}$$

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$$I = \int_0^{\pi/2} 3\sin^2 x\cos x dx = \left[\int 3\sin^2 x\cos x dx\right]_{x=0}^{x=\pi/2}$$

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$$\frac{du}{dx}=\frac{d}{dx}\sin x = \cos x \mbox{ $B$h$j(B } du = \cos x dx$$

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$$I = \left[\int 3u^2 du\right]_{x=0}^{x=\pi/2} = \left[u^3+C\right]_{x=0}^{x=\pi/2}$$

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$$I = \left[u^3+C\right]_{u=0}^{u=1} = (1+C)-(0-C) = 1$$

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$$I = \left[\sin^3 x+C\right]_{x=0}^{x=\pi/2} = \sin^3\frac{\pi}{2} -\sin^3 0=1$$

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$$I = \int_0^{\pi/2} x\cos x dx =\left[ \int x\cos x dx \right]_0^{\pi/2}$$

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$$\begin{eqnarray*}I & = & \left[\int x (\sin x)' dx\right]_0^{\pi/2} \\ & = & \... ...} + C\right) -(0\sin 0 + \cos 0 + C)\\ & = & \frac{\pi}{2}-1 \end{eqnarray*}$$

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