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evaluate the integral
- Thread starter jeca86
- Start date
There are different methods to solve, but one way is to let
\(\displaystyle x=4sin({\theta})\) and \(\displaystyle d{\theta}=4cos({\theta})\)
Remember, \(\displaystyle 1-sin^{2}({\theta})=cos^{2}({\theta})\)
\(\displaystyle \frac{64sin^{3}({\theta})}{sqrt{16(1-sin^{2}({\theta}))}}4cos({\theta})\)
\(\displaystyle 0=4sin({\theta}), {\theta}=0\)
\(\displaystyle 2\sqrt{3}=4sin({\theta}), {\theta}=\frac{\pi}{3}\)
This simplifies to:
\(\displaystyle 64\int_{0}^{\frac{{\pi}}{3}}{sin^{3}({\theta})d{\theta}\)
I initially forgot to change my parameters. I was bad. Entering the limits of integration into \(\displaystyle x=4sin({\theta})\)
Now, since \(\displaystyle sin^{2}({\theta})=1-cos^{2}({\theta})\), we have:
\(\displaystyle 64\int_{0}^{\frac{{\pi}}{3}}(1-cos^{2}({\theta}))sin({\theta})d{\theta}\)
Let \(\displaystyle u=cos({\theta})\) and \(\displaystyle du=-sin({\theta})d{\theta}\)
Change parameters again:
\(\displaystyle cos(0)=1, cos(\frac{{\pi}}{3})=\frac{1}{2}\)
Now integrate:
\(\displaystyle {-64}\int_{1}^{\frac{1}{2}}(1-u^{2})du\)
After you integrate and sub \(\displaystyle cos({\theta})\) back in,
remember that \(\displaystyle cos({\theta})=\frac{\sqrt{16-x^{2}}}{4}\)
That sure seems like a lot, doesn't it. Well, think of the Karate Kid painting Mr. Miyagi's fence and polishing the cars. It all benefited him in the end, though he didn't see it right off.
\(\displaystyle x=4sin({\theta})\) and \(\displaystyle d{\theta}=4cos({\theta})\)
Remember, \(\displaystyle 1-sin^{2}({\theta})=cos^{2}({\theta})\)
\(\displaystyle \frac{64sin^{3}({\theta})}{sqrt{16(1-sin^{2}({\theta}))}}4cos({\theta})\)
\(\displaystyle 0=4sin({\theta}), {\theta}=0\)
\(\displaystyle 2\sqrt{3}=4sin({\theta}), {\theta}=\frac{\pi}{3}\)
This simplifies to:
\(\displaystyle 64\int_{0}^{\frac{{\pi}}{3}}{sin^{3}({\theta})d{\theta}\)
I initially forgot to change my parameters. I was bad. Entering the limits of integration into \(\displaystyle x=4sin({\theta})\)
Now, since \(\displaystyle sin^{2}({\theta})=1-cos^{2}({\theta})\), we have:
\(\displaystyle 64\int_{0}^{\frac{{\pi}}{3}}(1-cos^{2}({\theta}))sin({\theta})d{\theta}\)
Let \(\displaystyle u=cos({\theta})\) and \(\displaystyle du=-sin({\theta})d{\theta}\)
Change parameters again:
\(\displaystyle cos(0)=1, cos(\frac{{\pi}}{3})=\frac{1}{2}\)
Now integrate:
\(\displaystyle {-64}\int_{1}^{\frac{1}{2}}(1-u^{2})du\)
After you integrate and sub \(\displaystyle cos({\theta})\) back in,
remember that \(\displaystyle cos({\theta})=\frac{\sqrt{16-x^{2}}}{4}\)
That sure seems like a lot, doesn't it. Well, think of the Karate Kid painting Mr. Miyagi's fence and polishing the cars. It all benefited him in the end, though he didn't see it right off.
Hello, jeca86!
Did you try any kind of substitution?
Substitute: \(\displaystyle \L\:\int\frac{x^2\cdot x}{\sqrt{16\,-\,x^2}}\,dx\;=\;\int\frac{(16\,-\,u)\cdot x}{u^{^{\frac{1}{2}}}}\cdot\left(-\frac{du}{2x}\right) \;= \;-\frac{1}{2}\int\frac{16\,-\,u}{u^{^{\frac{1}{2}}}}\,du\)
Now you can integrate: \(\displaystyle \L\:-\frac{1}{2}\int\left(16u^{^{-\frac{1}{2}}} \,- \,u^{^{\frac{1}{2}}}\right)\,du\) . . . and evaluate.
Did you try any kind of substitution?
Let: \(\displaystyle u\:=\:16\,-\,x^2\;\;\Rightarrow\;\;x^2\:=\:16\,-\,u\;\;\Rightarrow\;\;2x\,dx\,=\,-du\;\;\Rightarrow\;\;dx\,=\,-\frac{du}{2x}\)Evaluate the integral: \(\displaystyle \L\:\int^{\;\;\;2\sqrt{3}}_0\frac{x^3}{\sqrt{16\,-\,x^2}}\,dx\)
Substitute: \(\displaystyle \L\:\int\frac{x^2\cdot x}{\sqrt{16\,-\,x^2}}\,dx\;=\;\int\frac{(16\,-\,u)\cdot x}{u^{^{\frac{1}{2}}}}\cdot\left(-\frac{du}{2x}\right) \;= \;-\frac{1}{2}\int\frac{16\,-\,u}{u^{^{\frac{1}{2}}}}\,du\)
Now you can integrate: \(\displaystyle \L\:-\frac{1}{2}\int\left(16u^{^{-\frac{1}{2}}} \,- \,u^{^{\frac{1}{2}}}\right)\,du\) . . . and evaluate.