- #1

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

Thank you..

- Thread starter Mag|cK
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- #1

- 36

- 0

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

Thank you..

- #2

cristo

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

- #3

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yes i have done using substitution method but can't solve the problem. Can you step by step show me how please? thx^{2}(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)

- #4

cristo

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

- #5

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

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 cant eliminate the cos5x.

I already tried substituting with cosec5x and sec5x, but still no solution.

- #6

jakcn001

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

- #7

cristo

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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-uok 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 cant eliminate the cos5x.

There you go, I've done one step for you. Now, try the substitution v=1/u.

If you look back, I said you'll need aI already tried substituting with cosec5x and sec5x, but still no solution.

I did allude to that in my first post.With -d[cot(x)]=(sinx)^(-2) dx ,you can solve it easily.Have a try!

- #8

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ok thx guys i understand now

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