Find the integral $\int_{}\sec^3x$ via u-substitution
I already know that there are multiple answers of integrating $\int_{}\sec^3x$. My textbook uses an integration by parts approach and I also have seen a partial functions approach as well.
I stumbled upon this post and one of the users @N.S. posted an interesting approach that used substitution, but I wasn't able to work the problem out completely...Here is where I got to
$$\int_{} \sec^3(x) = \int \frac{1}{\cos^3x} * \frac{\cos(x)}{\cos(x)} = \int \frac{\cos(x)}{\cos^4(x)} = \int \frac{\cos(x)}{(1- \sin^2(x))^2}dx $$
Apply substitution with $u = \sin(x), du = \cos(x)$:
$$\int \frac{du}{(1-u^2)^2} = \int \frac{du}{1 - 2u^2 + u^4} = \frac{1}{\frac{2}{3}u^3 - \frac{u^5}{5}} $$
But this just seems wrong and nowhere close to the answer of $\frac{1}{2} (\sec(x) \tan(x) + ln| \sec(x) + \tan (x)|) + C$
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$\begingroup$How did you perform that final integration? It looks like you've assumed some nonexistent identity. Go with partial fractions (for $\frac{1}{1-u^2}$ first, then square it). You'll find $(\sec x+\tan x)^2=\frac{1+\sin x}{1-\sin x}$ useful.
$\endgroup$ 1 $\begingroup$Keep one cosine, and convert the others to sine. Then substutute $u = \sin x, du = \cos x\;dx$.\begin{align} \int \sec^3 x\;dx &= \int\frac{\cos x\;dx}{\cos^4 x} = \int\frac{\cos x\;dx}{\left(1-\sin^2 x\right)^2} \\&= \int \frac{du}{(1-u^2)^2} \end{align}so it is reduced to a rational function. All that was correct. But you did not integrate the rational function correctly. Perhaps use the method of partial fractions?$$ \frac{1}{(1-u)^2(1+u)^2} = \frac{1}{4(1+u)}+\frac{1}{4(1+u)^2} +\frac{1}{4(1-u)}+\frac{1}{4(1-u)^2} $$
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