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Integration by parts possibly: int( sqrt(36-x^2)/x)dx
- Thread starter jman2807
- Start date
Re: Integration by parts possibly
Hello, jman2807!
Mark is absolutely correct: try trig substitution.
Let \(\displaystyle x\,=\,6\cdot\sin\theta\;\;\Rightarrow\;\;dx\,=\,6\cdot\cos\theta\cdot d\theta\)
. . and the radical becomes: \(\displaystyle \,6\cdot\cos\theta\)
Substitute: \(\displaystyle \L\:\int \frac{6\cdot\cos\theta}{6\cdot\sin\theta}\)\(\displaystyle \,(6\cdot\cos\theta\cdot d\theta) \;=\;6\L\int\frac{\cos^2\theta}{\sin\theta}\)\(\displaystyle \,d\theta \;=\;6\L\int\frac{1\,-\,\sin^2\theta}{\sin\theta}\)\(\displaystyle \,d\theta\)
. . \(\displaystyle = \;6\L\int\left(\frac{1}{\sin\theta} \,-\,\frac{\sin^2\theta}{\sin\theta}\right)\)\(\displaystyle \,d\theta \;= \;6\L\int\)\(\displaystyle (\csc\theta\,-\,\sin\theta)\,d\theta\)
Can you finish it now?
Hello, jman2807!
Mark is absolutely correct: try trig substitution.
\(\displaystyle \L\int \frac{\sqrt{36\,-\,x^2}}{x}\,dx\)
Let \(\displaystyle x\,=\,6\cdot\sin\theta\;\;\Rightarrow\;\;dx\,=\,6\cdot\cos\theta\cdot d\theta\)
. . and the radical becomes: \(\displaystyle \,6\cdot\cos\theta\)
Substitute: \(\displaystyle \L\:\int \frac{6\cdot\cos\theta}{6\cdot\sin\theta}\)\(\displaystyle \,(6\cdot\cos\theta\cdot d\theta) \;=\;6\L\int\frac{\cos^2\theta}{\sin\theta}\)\(\displaystyle \,d\theta \;=\;6\L\int\frac{1\,-\,\sin^2\theta}{\sin\theta}\)\(\displaystyle \,d\theta\)
. . \(\displaystyle = \;6\L\int\left(\frac{1}{\sin\theta} \,-\,\frac{\sin^2\theta}{\sin\theta}\right)\)\(\displaystyle \,d\theta \;= \;6\L\int\)\(\displaystyle (\csc\theta\,-\,\sin\theta)\,d\theta\)
Can you finish it now?
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
Re: Integration by parts possibly
substitute: x = 6 sin(u)
jman2807 said:I have a problem which i think may be integration by parts but i cant seem to get a start on it.
int( sqrt(36-x^2)/x)
sorry for the lack of tex hopefully you can read. Thanks in advance.
substitute: x = 6 sin(u)