- #1
bomba923
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(excuse my poor LaTex...i don't know it very well yet 
)
2\pi \int {x^3 \sin 2x\,dx} \Rightarrow \left\{ \begin{array}{l}<br /> u = 2x \\ <br /> du = 2dx \\ <br /> \end{array} \right\} \Rightarrow \frac{\pi }{8}\int {u^3 \sin u\,du} = \frac{\pi }{8}\left( { - u^3 \cos u + \int {u^2 \cos u\,du} } \right)
= \frac{\pi }{8}\left( { - u^3 \cos u + u^2 \sin u + u\cos u - \sin u} \right)
= \frac{{\pi \left[ { - 8x^3 \cos \left( {2x} \right) + 4x^2 \sin \left( {2x} \right) + 2x\cos {2x} - \sin {2x}} \right]}}{8}
How can I do this faster? Are there things I can skip or connect--etc--?
)2\pi \int {x^3 \sin 2x\,dx} \Rightarrow \left\{ \begin{array}{l}<br /> u = 2x \\ <br /> du = 2dx \\ <br /> \end{array} \right\} \Rightarrow \frac{\pi }{8}\int {u^3 \sin u\,du} = \frac{\pi }{8}\left( { - u^3 \cos u + \int {u^2 \cos u\,du} } \right)
= \frac{\pi }{8}\left( { - u^3 \cos u + u^2 \sin u + u\cos u - \sin u} \right)
= \frac{{\pi \left[ { - 8x^3 \cos \left( {2x} \right) + 4x^2 \sin \left( {2x} \right) + 2x\cos {2x} - \sin {2x}} \right]}}{8}
How can I do this faster? Are there things I can skip or connect--etc--?
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