Question

How do I use trig substitution to integrate:

sqrt(25 - x^2)/x?

I know that I start with x = 5sin(theta) but I am not able to solve it

Answer 1

Please show your steps so I can follow

Answer 2

Well, I hope I got what you meant right..
Look at attachment.
We know that
\[
\frac{AC}{BC} = \frac{\sqrt{25 - x^2}}{x} = cot(\theta)
\]

Answer 3

Yes but then I get integral of cot(theta)cos(theta) d(theta)... I don't know where to go from there

Answer 4

because dx = cos(theta) d(theta)

Answer 5

\[I=\int\limits \frac{ \sqrt{25-x ^{2}} }{ x }dx\]
\[put x=5 \sin \theta ,dx=5\cos \theta d \theta \]
\[I=\int\limits \frac{ \sqrt{25-25\sin ^{2}\theta } }{ 5\sin \theta } 5\cos \theta d \theta \]
\[=\int\limits \frac{ 5\cos ^{2}\theta }{\sin \theta }d \theta =5\int\limits \frac{ \cos ^{2}\theta \sin \theta d \theta }{ \sin ^{2}\theta }\]
\[=5\int\limits \frac{ \cos ^{2}\theta \sin \theta d \theta }{1-\cos ^{2}\theta }\]
\[put \cos \theta =t,-\sin \theta d \theta=dt\]
\[I=5\int\limits \frac{ -t ^{2}dt }{1-t ^{2} }=5\int\limits \frac{ 1-t ^{2}-1 }{ 1-t ^{2} }dt\]
=\[=5\int\limits1dt-\int\limits \frac{ 1 }{1-t ^{2} }dt=5t-I1 ......(1)\]
\[I1=\int\limits \frac{ 5 }{ 1-t ^{2} }dt\]
\[\frac{ 5 }{ 1-t ^{2} }=\frac{ 5 }{ \left( 1-t \right)\left( 1+t \right) }=\frac{ 5 }{\left( 1-t \right)\left( 1+t \right) }\]
\[=\frac{ 5 }{\left( 1-t \right)\left( 1+1 \right) }+\frac{ 5 }{ \left( 1+1 \right)\left( 1+t \right) }\]
\[=\frac{ 5 }{2 } \frac{ 1 }{1-t }+\frac{ 5 }{2 }\frac{ 1 }{ 1+t }\]
\[I1=\frac{ 5 }{2 }\int\limits \frac{ 1 }{1-t }dt+\frac{ 5 }{ 2 }\int\limits \frac{ dt }{1+t }\]
\[I1=-\frac{ 5 }{ 2 }\ln \left( 1-t \right)+\frac{ 5 }{2 }\ln \left( 1+t \right)\]
\[I=5t+\frac{ 5 }{ 2 }\ln \frac{ 1-t }{1+t }+c\]
replace the values of t and then cos theta and sin theta and get the answer.

Answer 6

correction in the line no. (1) 5 is missing twice.

Answer 7

have you followed or should I complete.

Answer 8

thanks I got it already but now I know I did the right thing