Possible webpage title: Solving Integrals Using the Substitution Method

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The forum discussion focuses on solving integrals using the substitution method, specifically evaluating the integral ∫1/(1-sin²u)×cosu du. The user proposes the substitution x = sinu, leading to the integral being simplified to ∫1 du = u + c = sin⁻¹x + c. Another participant suggests an alternative substitution, u = 1 - x², which may simplify the process further. The discussion emphasizes the effectiveness of substitution methods in integral calculus.

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Evaluate the following integral using integration by substitution: http://img254.imageshack.us/img254/750/44900023cm4.png [/URL]




Here is my attempt:
Let x = sinu, then dx/du = cosu
Substituting gives, ∫1/(1-sin2u)×cosu du
= ∫1/(1-sin2u)×cosu du
= ∫cosu/√cos2u du
= ∫cosu/cosu du
= ∫1 du = u + c = sin-1x + c

Am I right? Did I get the right solution?

Regards,

 
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You don't need Trig sub. for this.

[tex]\int\frac{xdx}{\sqrt{1-x^2}}[/tex]

But we'll go with it!

[tex]x=\sin u[/tex]
[tex]dx=\cos udu[/tex]

[tex]\int\frac{\sin u \cos u du}{\sqrt{1-\sin^2 u}}[/tex]
 
Try u=1-x^2 instead... Might be a bit easier.
 

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