<s>[INST] Integration on $ \int \sec^3 (2x) dx$ The answer is this but the coefficient is $\frac{1}{4}$. Why? 
My steps:
$$\int \sec^3(2x)dx$$ 
Let $u = 2x$, then $\frac{1}{2}du = dx$
$$\frac{1}{2} \int \sec^3(u) du $$ 
Using int my parts: Let $a = \sec(u)$, then $da = \sec(u)\tan(u)du$. let $dv = \sec^2 (u) du$, then $v = \tan(u)$
$$\frac{1}{2} \int \sec^3(u) du = \frac{1}{2}\big(\sec(u)\tan(u) - \int \sec(u)\tan^2(u)du\big)$$
$$= \frac{1}{2}\big(\tan(u)\sec(u) - \int \sec^3 (u)du + \int \sec(u)du\big)$$
$$\frac{1}{2} \int \sec^3(u)du + \frac{1}{2} \int \sec^3(u)du = \frac{1}{2}\big(\tan(2x)\sec(2x) + \ln |\sec(2x)+\tan(2x)|\big) + C$$
$$\int \sec^3(2x)dx = \frac{1}{2}\big(\tan(2x)\sec(2x) + \ln |\sec(2x)+\tan(2x)|\big) + C$$
 [/INST] What you derived is that $\int \sec^3 (u)~du=\text{blah}$. Then note that $\int \sec^3(2x)~dx=\frac12\int \sec^3(u)~du$
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