Integral involving square root...
- Thread starter Lizzy
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skeeter
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let \(\displaystyle \L u = 2 - x^3\)
\(\displaystyle \L du = -3x^2 dx\)
\(\displaystyle \L x^3 = 2 - u\)
\(\displaystyle \L \int x^5 \sqrt{2 - x^3} dx = -\frac{1}{3} \int x^3 \cdot (-3x^2) \sqrt{2 - x^3} dx\)
substitute ...
\(\displaystyle \L -\frac{1}{3} \int (2 - u) \sqrt{u} du\)
\(\displaystyle \L \frac{1}{3} \int u^{\frac{3}{2}} - 2u^{\frac{1}{2}} du\)
integrate and back substitute
\(\displaystyle \L du = -3x^2 dx\)
\(\displaystyle \L x^3 = 2 - u\)
\(\displaystyle \L \int x^5 \sqrt{2 - x^3} dx = -\frac{1}{3} \int x^3 \cdot (-3x^2) \sqrt{2 - x^3} dx\)
substitute ...
\(\displaystyle \L -\frac{1}{3} \int (2 - u) \sqrt{u} du\)
\(\displaystyle \L \frac{1}{3} \int u^{\frac{3}{2}} - 2u^{\frac{1}{2}} du\)
integrate and back substitute