Need help on integral question

[Mag|cK]

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..

 

[cristo]

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)

 

[Mag|cK]

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

 

[cristo]

It's your homework, not mine! Show how you did the first substitution, and we'll go from there.

 

[Mag|cK]

It's your homework, not mine! Show how you did the first substitution, and we'll go from there.

ok sorry
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.

 

[jakcn001]

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

 

[cristo]

ok sorry
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!

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

I did allude to that in my first post.

 

[Mag|cK]

ok thanks guys i understand now