Solving Trig Anti-Derivative: Int of sin x over -sin^2 x of dx

[jwxie]

Homework Statement

Int of sin x over - sin ^2 x of dx

The Attempt at a Solution

I don;t know if I have the right question. But I just couldn't reproduce the problem (from my answer)!
please take a look at my note, see if you can reproduce the original question

i am talking about question #7

 

[Mark44]

Your attachment is still pending approval.
Is this your integral?
$$\int \frac{sin x}{-sin^2 x}dx$$

If so, this is the same as
$$\int -csc (x) dx$$
If I recall correctly, this can be done using integration by parts.

 

[jwxie]

yeah i think that;s what it is
but if we use -csc x can it be done easily?
i mean cscx does not have any Anti-derv...

 

[Mark44]

But -sin^2(x) isn't equal to cos^2(x), so your first step is incorrect.

 

[jwxie]

then shouldn't it be int of sinx times -sin^2 x?

 

[Mark44]

then shouldn't it be int of sinx times -sin^2 x?

I don't know what you mean. This is what you wrote in your first post in this thread:

Int of sin x over - sin ^2 x of dx

By "over" I assume you mean the quotient of sin(x) and -sin^2(x), which is what I showed in the integral.

 

[jwxie]

i am sorry, i am referring to your #4
you said my first step was wrong

now, i knew the mistake, and this is what i did

original question:
Int of sin x over - sin ^2 x of dx

first step, change the bottom, -sin^2 (x) to this form --> 1 over csc^2 (x)

so the entire int will become

int of sin (x) times csc^2 (x)

because 1/sinx = cscx, then 1 / -sin^2 (x) = csc^2 (x)
am i correct?

 

[Mark44]

i am sorry, i am referring to your #4
you said my first step was wrong

now, i knew the mistake, and this is what i did

original question:
Int of sin x over - sin ^2 x of dx

Or, using inline LaTeX tags,
$\int sin(x)/(-sin^2(x)) dx$

first step, change the bottom, -sin^2 (x) to this form --> 1 over csc^2 (x)

You've lost a sign. -sin^2(x) = -1/csc^2(x)

so the entire int will become

int of sin (x) times csc^2 (x)

Or $\int sin(x)(-csc^2(x)) dx$

because 1/sinx = cscx, then 1 / -sin^2 (x) = csc^2 (x)
am i correct?

No.
1 / -sin^2 (x) = -csc^2 (x)

 

[Brian_C]

You're going around in circles. The integrand simplifies to -1/sin x or -csc x.

It is not at all obvious how to integrate csc x. Try multiplying the integrand by a certain factor which allows you to make a substitution, but which doesn't change the value of the integrand. Hint: the final result involves a logarithm.