$\displaystyle \int 3\mathrm{sec}^3x\tan x \ \mathrm{d}x$ Thanks in advance.
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$\displaystyle \int 3\sec^3x\tan xdx=3\int \frac{\sin x}{\cos^4x}dx$. Use the substitution $\displaystyle \cos x=u$
Originally Posted by Air $\displaystyle \int 3\mathrm{sec}^3x\tan x \ \mathrm{d}x$ Thanks in advance. Alternatively we have $\displaystyle 3\int\sec^2(x)\cdot\sec(x)\tan(x)dx$ Let $\displaystyle u=\sec(x)\Rightarrow{du=\sec(x)\tan(x)}$ Giving us $\displaystyle 3\int{u^2du}$
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