How do I solve this integrals?(dy/secˆ2(y))(dx/(-xˆ2 + x))

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SUMMARY

The discussion focuses on solving the integrals (dy/sec²(y)) and (dx/(-x² + x)). The first integral can be simplified using the identity ∫(dy/sec²(y)) = ∫cos²(y) dy, leveraging the double-angle formula. For the second integral, participants suggest completing the square in the denominator and applying u-substitution to facilitate the integration process.

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  • Knowledge of u-substitution techniques in integration
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How do I solve this integrals?

(dy/secˆ2(y))

(dx/(-xˆ2 + x))
 
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FrostScYthe said:
How do I solve this integrals?

(dy/secˆ2(y))

Partial integration or use the formula's linking cos(x) to cos(2x)

(dx/(-xˆ2 + x))

Partial fractions .

marlon
 
Last edited:
FrostScYthe said:
How do I solve this integrals?

(dy/secˆ2(y))

(dx/(-xˆ2 + x))
#1, [tex]\int \frac{dy}{\sec ^ 2 y} = \int \frac{dy}{\frac{1}{\cos ^ 2 y}} = \int \cos ^ 2 y dy[/tex]
Now, do you know the Double-angle formulae?
#2, I'll give you a hint, try to complete the square in the denominator, then use u-substitution, and it's done:
[tex]\int \frac{dx}{-x ^ 2 + x} = - \int \frac{dx}{x ^ 2 - x}[/tex]
Can you go from here? :)
 

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