How Do You Solve the Integral of (sin(5x))^(-2) dx?

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Hello, i have a question. This is integral question, but i don't have any program to write the equation with:

integral (sin5x)^(-2) dx= ?

Thank you..
 
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This is one of those you should really know, since it can be rewritten cosec2(5x).

However, if you don't know the antiderivative, then it can be worked out by a few substitutions, the first being u=sin(5x)
 
cristo said:
This is one of those you should really know, since it can be rewritten cosec2(5x).

However, if you don't know the antiderivative, then it can be worked out by a few substitutions, the first being u=sin(5x)

yes i have done using substitution method but can't solve the problem. Can you step by step show me how please? thx
 
It's your homework, not mine! Show how you did the first substitution, and we'll go from there.
 
cristo said:
It's your homework, not mine! Show how you did the first substitution, and we'll go from there.

ok sorry :smile:
if i substitute with sin5x,

integral (sin5x)^(-2) d(sin5x)/5cos5x

this is how far i can get if i substitute with sin5x as i can't eliminate the cos5x.
I already tried substituting with cosec5x and sec5x, but still no solution.
 
With -d[cot(x)]=(sinx)^(-2) dx ,you can solve it easily.Have a try!
 
Mag|cK said:
ok sorry :smile:
if i substitute with sin5x,

integral (sin5x)^(-2) d(sin5x)/5cos5x

this is how far i can get if i substitute with sin5x as i can't eliminate the cos5x.
If you've substituted, why do you still have functions of x in the integral? If you let u=sin(5x), then du=5cos(5x)dx=5[sqrt(1-u2)]. This transforms your integral to 5 \int \frac{du}{u^2\sqrt{1-u^2}}.

There you go, I've done one step for you. Now, try the substitution v=1/u.
I already tried substituting with cosec5x and sec5x, but still no solution.
If you look back, I said you'll need a few substitutions!


jakcn001 said:
With -d[cot(x)]=(sinx)^(-2) dx ,you can solve it easily.Have a try!

I did allude to that in my first post.
 
ok thanks guys i understand now :smile:
 
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