Best Answer:
I consider the first subject is the integral for t. Because your question is about that sec-1t.

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If you have else questions or find some mistakes which I write. Please tell me.

2008-01-15 21:51:39 補充：

Some words are covered. The content of the words are:

I consider the first subject is the integral for t. Because your question is about that arcsect.

2008-01-15 21:57:46 補充：

Some words are covered. The content of the words is:

I consider the first subject is the integral for t. Because your question is about that arcsect.

2008-01-15 22:34:35 補充：

What you get is 1/[x^3*根號(t^2-1)] , there are two conditions, for x and for t.

If your first subject is∫1/[t^3*(x^2-1)]dt, you can use my solution directly.

If it is∫1/[x^3*(x^2-1)]dx and let x=sect. Let θconsider t.

2008-01-15 22:34:42 補充：

Your result:1/2*arcsen(t)+根號(t^2-1)/2t^2

I can't calculate the result. Maybe you need to check your algorithm.

2008-01-16 01:00:07 補充：

I cheak again. I find that I have some calculation mistakes.

Sorry! Your result is right.

I seem to let sin2θ=sinθcosθ, I should multiply 2.

2008-01-16 01:00:18 補充：

In fact, you don't need to write the complex esult. In the trigonometric substitution, when we let t=sinθ, we don't need to change fromθ to t again. Because you also change the upper and lower limit. It's too troublesome.

2008-01-16 10:49:09 補充：

arcsect:[0,π/2)→ the value is positive. [π,3π/2)→negative

(arcsec2)/2+[(2^2-1)^1/2]/2*(2^2)-{arcs...

=(π/3)/2+(√3)/8-(π/4)/2-1/4=π/24+(√3)/8...

About 1st subjest, do you understand?

Source(s):
Me

Anonymous
· 1 decade ago