trigonometry substitution

**Archduke01** · March 24th 2010, 06:00 PM

$\int \frac {1}{x^2sqrt(1-x^2)} dx$

$x = sin t$
$dx = cos t dt$

$= \int \frac{cos t}{sin^2tsqrt(1-sin^2t)}$

$= \int \frac {cos t}{sin^2tsqrt(cos^2t)}$

$= \int \frac {1}{sin^2t}$

That's as far as I got ... how can I proceed?

PS; I apologize for the untidiness of the square roots. How do I make the radical appear?

**ione** · March 24th 2010, 06:06 PM

Originally Posted by Archduke01

$\int \frac {1}{x^2sqrt(1-x^2)} dx$

$x = sin t$
$dx = cos t dt$

$= \int \frac{cos t}{sin^2tsqrt(1-sin^2t)}$

$= \int \frac {cos t}{sin^2tsqrt(cos^2t)}$

$= \int \frac {1}{sin^2t}$

That's as far as I got ... how can I proceed?

PS; I apologize for the untidiness of the square roots. How do I make the radical appear?

$\frac {1}{sin^2t}=csc^2t$

**Archduke01** · March 24th 2010, 06:22 PM

OK, so I finally end up with $- cot t + C$

I'm confused with making the triangle though. I have my $x = sin t$ , but using that, all my sides will basically be 1. Is that right?

**ione** · March 24th 2010, 06:34 PM

Originally Posted by Archduke01

I'm confused with making the triangle though. I have my $x = sin t$ , but using that, all my sides will basically be 1. Is that right?

t is an acute angle of the triangle. The side opposite t is x, the hypotenuse is 1, and the side adjacent to t is $\sqrt{1-x^2}$

**drumist** · March 24th 2010, 06:43 PM

Are you evaluating this over an interval or is it an indefinite integral?

The method you use is a bit tricky if it is not over a small interval (i.e., one where cos x is always positive or negative) because:

$\frac{\cos x}{\sqrt{\cos^2 x}} = \frac{\cos x}{|\cos x|} = \begin{cases} 1, & \cos x > 0 \\ -1, & \cos x < 0\end{cases}$

P.S. Mouse over the math symbols to see the code used to generate them.

You could also do this problem using $u=\sqrt{1-x^2}$

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