How do you integrate $\int {sec}^{6} (4 x) tan (4 x)$ $int sec^6(4x)tan(4x)$ ?

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Answer 1 · 2016-12-08T04:50:02

$\frac{{sec}^{6} (4 x)}{24} + C$ $sec^6(4x)/24+C$

Use dsecx = sec x tan x dx.

$\int {sec}^{6} (4 x) tan (4 x) d x$ $int sec^6(4x)tan(4x) dx$

$= \frac{1}{4} \int {sec}^{5} (4 x) d sec (4 x)$ $=1/4 int sec^5(4x)dsec(4x)$

$= \frac{{sec}^{6} (4 x)}{24} + C$ $=sec^6(4x)/24+C$

How do you integrate int sec^6(4x)tan(4x)∫sec6(4x)tan(4x)int sec^6(4x)tan(4x)?