How do you find the integral of $\int {tan}^{3} (2 x) {sec}^{5} (2 x) d x$ $int tan^3(2x) sec^5(2x) dx$ ?

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Answer 1 · 2015-10-17T14:05:11

$I = \frac{1}{2} [\frac{{sec}^{7} 2 x}{7} - \frac{{sec}^{5} 2 x}{5}] + C$ $I = 1/2 [(sec^7 2x)/7-(sec^5 2x)/5] + C$

$sec2x=t => sec2xtan2x*(2x)'dx=dt$

$2sec2xtan2xdx=dt$

$sec^2 2x=1+tan^2 2x => tan^2 2x = sec^2 2x -1$

$I = int tan^3 2x sec^5 2x dx$

$I = int tan^2 2x sec^4 2x tan2x sec 2x dx$

$I = 1/2 int (sec^2 2x -1) sec^4 2x (2tan2x sec 2x dx)$

$I = 1/2 int (t^2 -1) t^4 dt = 1/2 int (t^6 - t^4) dt = 1/2 [t^7/7-t^5/5] + C$

$I = 1/2 [(sec^7 2x)/7-(sec^5 2x)/5] + C$

How do you find the integral of int tan^3(2x) sec^5(2x) dx∫tan3(2x)sec5(2x)dxint tan^3(2x) sec^5(2x) dx?