Self-learning calculus - Integral [Sin(-1) x] dx

[jackson6612]

I'm self-learning calculus. I'm not good at maths. Well, I just tried to solve a random problem in a math software

Integral [ Sin^(-1) x ] dx

The solution is here:
http://img43.imageshack.us/img43/4597/60496310.png

Now the problem is I don't how it's going. How do we get to step 2 involing t's?

Please help me. I would be highly indebted to you for all the guidance. Thanks.

 

[eumyang]

Use the trig identity
\sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta}
with θ = x/2. Then substitute t = tan(x/2).

 

[jackson6612]

Thanks, Yang.

But what does the given formula have to do with [Sin(-1) x]. We know the ratio, x, and need to know the angle.

Sin(theta) = x, where 'x' is a ratio and 'theta' is an angle. In Sin(-1) situation has been reversed, now we know 'x', not 'theta'. Am I correct?

Please guide me.

 

[micromass]

Oh, you talking about the inverse of the sine function... This is usually denoted as arcsin.

Have you tried partial integration?

 

[jackson6612]

No, I'm new to this calculus world. Please have a look on the link in my first post. Is it really arcsin? Thanks.

 

[micromass]

Sorry, didn't see the link.

eumyang gave some good advice: just substitute t=tan(x/2)...

 

[Mark44]

Based on what I see in the OP's imageshack image, this is the integral:
\int \frac{dx}{sin(x)}

The OP's notation of sin^(-1)(x) was very misleading, as that notation is used for the inverse sine function.

 

[jackson6612]

Sorry, didn't see the link.

eumyang gave some good advice: just substitute t=tan(x/2)...

Micro, I'm the one to be blamed. Sorry.

 

[jackson6612]

Based on what I see in the OP's imageshack image, this is the integral:
\int \frac{dx}{sin(x)}

The OP's notation of sin^(-1)(x) was very misleading, as that notation is used for the inverse sine function.

Mark, thanks a lot. Now I could at least where it's coming from. Will get back.

 

[Mark44]

There substitution of t = tan(x/2) is not, IMO, an obvious one. Here's another way to do the problem.
\int \frac{dx}{sin(x)} = \int csc(x) dx
=\int csc(x) \frac{csc(x) + cot(x)}{csc(x) + cot(x)}dx
=\int \frac{csc^2(x) + cot(x)csc(x)}{csc(x) + cot(x)}dx

Now, you can use the substitution u = csc(x) + cot(x), so du = -csc(x)cot(x) - cot²(x) dx, so the integral above can be rewritten as
\int \frac{-du}{u} = -ln|u| + C = -ln|csc(x) + cot(x)| + C

Although this result looks very different from the answer in the imageshack link (ln|tan(x/2)| + C), the two expressions are equal.

 

[jackson6612]

In the first step where does the factor 2/(1+t^2) come from? Please help me. Thanks.

 

[jackson6612]

Mark, I didn't see your post until I was done with mine. Thanks a lot for the alternative. I understand it takes a lot of effort and time. But for some reason the alternative is somewhat fearful for me. My shortcomings are to be blamed here. I genuinely thank you for your input. Best wishes.

 

[Mark44]

They seem to be using right triangle trig based on the substitution t = tan(x/2). It's helpful to draw a right triangle with base 1 and height t, and acute angle x. If you extend the base to 2 units, the new triangle has a base of 2 units and height of t (still), but now the acute angle is x/2. From those two triangles you can get the various relationships between t and x or x/2.

 

[jackson6612]

Mark: I just had another look at your method and now it seems it's not difficult. Now your effort has been fully utilized, I believe. Thanks a lot.

BTW, how do you write definite integral in Latex? What's the code? Could you please tell me?

 

[micromass]

Click with your mouse on the LaTeX to see the code:

\int{\frac{1}{x}dx}

 

[jackson6612]

Hi Micro

You have missed it: I said "definite integral". I suppose that's what it's called - an integral sign with two number, one below it and other above it.

 

[micromass]

Oops, sorry... Here's the right code:

\int_{-2}^3{\frac{1}{x}dx}

So in short: just append _{lower bound}^{upper bound} to the \int.

 

[jackson6612]

Thank you.

 

[HallsofIvy]

Use "_" for the lower limit and "^" for the upper
\int_a^b f(x)dx

is \int_a^b f(x)dx

 

[jackson6612]

Thank you, Halls.

 

[Mark44]

Also, if either of the limits of integration has more than one character, put braces ( { and }) around the expression. This is true also for exponents.
\int_{-1}^{13} e^{2x}dx

The script inside the tex tags for the above is \int_{-1}^{13} e^{2x}dx

If you don't use braces, this is what happens:
\int_-1^13 e^2xdx

 

[jackson6612]

Mark: thanks for letting me know. It would take some time to be good at Latex.

 

[jackson6612]

In the first step where does the factor 2/(1+t^2)dt come from? Please help me. Thanks.

Please check the link in my first post in this thread to see the first step. Thanks.

 

[jackson6612]

Help, please!

 

[micromass]

It comes from the dx. If t=tan(x/2), dx=2dt/(1+t^2). So it's from the substitution formula.

 

[Bohrok]

There's another way that doesn't use non-intuitive substitutions like u = secx + tanx or u = tan(x/2). This shows how to find ∫secx dx, but integrating cscx is almost exactly the same.
http://www.karlscalculus.org/calc11_6.html

 

[jackson6612]

Hi Bohrok

Where did you get that integral symbol, ∫secx dx, from? I could even copy it to wordpad. Whenever I try to copy Latex written text, I get everything copied in Latex code. Please let me know. Thanks.