Trig Substitution With an x^3 in the Numerator

by Anonymous
(US)

So far I have x=secθ and dx=secθtanθdθ and substituted it in the equation for x and dx. I am now stuck at integral of ∫sec^4(θ) dθ. I'm not sure if I did everything up to that point correctly.

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Answer by Pablo:

You did excelent so far. For those that maybe didn't undertand what you did I'll rewrite it. In this case we use the substitution:



So, we have that:



And our integral becomes:



We now use the following trigonometric identity:



Applying that we get:



Now we're left with a trigonometric integral. To solve this one we apply the same trigonometric identity:



The firt integral is solved directly since the derivative of tangent is secant square. The second one requires a substitution:



Simply solving the first one and making the substitution in the second:



Now we only need to substitute everything back. We have that:



And that:



So our integral is:



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