Help with trigonometric substitutions

[afcwestwarrior]

Homework Statement

∫1/x^2*√25-x^2

Homework Equations

√a^2-x^2 , x=a sin (theta) , -pi/2 less than or equal theta less than or equal pi/2 Identity 1-sin^2 theta= cos^2 theta

table of trigonometric substitutions

The Attempt at a Solution

x=5 sin theta
dx= 5 cos theta dtheta
theta=arcsin (x/5)
√25-x^2=5 cos theta
∫5 cos theta / 25 sin^2 theta * (5cos theta)=1/25 ∫sin^2 theta d theta

=1/25 - cos^2 theta+c
= -1/25 cos^2 (arcsin(x/5) +c

but the answer in the back of the book is -√25-(x^2)/(25x) +c

what did i do wrong

 

[afcwestwarrior]

anyone know what i did wrong

 

[dynamicsolo]

Is the integral

\int \frac{dx}{x^2 \cdot \surd(25-x^2)}?

The Attempt at a Solution

x=5 sin theta
dx= 5 cos theta dtheta
theta=arcsin (x/5)
?25-x^2=5 cos theta
?5 cos theta / 25 sin^2 theta * (5cos theta)=1/25 ?sin^2 theta d theta

How did sin^2 suddenly get into the numerator? I believe the trig-substituted integral at this point will be

\frac{1}{25} \int \frac{d\theta}{sin^2 \theta}

Do we have an antiderivative handy for csc^2 \theta?

BTW, please don't "bump" your own threads: it makes the reply count go up, so it looks like someone has started helping you when they actually haven't yet. Please be patient and someone will get to you. It's going to be slow right now because most universities are still on "interim" until at least next week (and many don't start up again until well into September)...

 

[HallsofIvy]

Homework Statement

∫1/x^2*√25-x^2

First question: is this
\int \frac{dx}{x^2\sqrt{25- x^2}}
or
\int \frac{1}{x^2} \sqrt{25- x^2}dx?

Homework Equations

√a^2-x^2 , x=a sin (theta) , -pi/2 less than or equal theta less than or equal pi/2 Identity 1-sin^2 theta= cos^2 theta

table of trigonometric substitutions

The Attempt at a Solution

x=5 sin theta
dx= 5 cos theta dtheta
theta=arcsin (x/5)
√25-x^2=5 cos theta
∫5 cos theta / 25 sin^2 theta * (5cos theta)=1/25 ∫sin^2 theta d theta

=1/25 - cos^2 theta+c
= -1/25 cos^2 (arcsin(x/5) +c

but the answer in the back of the book is -√25-(x^2)/(25x) +c

what did i do wrong

If the problem is
\int \frac{dx}{x^2\sqrt{25- x^2}}
then the substitution x= 5 sin(t) gives dx= 5 cos(t)dt, \sqrt{25- x^2}= 5cos(t) and x²= 25 sin²(t) so the integral becomes
\int \frac{5 cos(t)dt}{(25 sin^2(t))(5cos(t))}= \frac{1}{25}\int \frac{dt}{sin^2(t)}= \frac{1}{25}\int csc^2(t) dt

If the problem is
int \frac{1}{x^2}\sqrt{25- x^2}dx
then the substitution x= 5 sin(t) gives
\int \frac{25 cos^2(t) dt}{25 sin^2(t)}= \int cot^2(t) dt[/itex]

 

[afcwestwarrior]

It's the first one

 

[dynamicsolo]

Then the replies in post #3 and the first integration in post #4 are appropriate. Find the general antiderivative of <br /> csc^2 \theta<br />, then back-substitute to get the expression in terms of x.

 

[afcwestwarrior]

thanks