What is the integral of $\int {tan}^{5} (x) \cdot {sec}^{4} (x) d x$ $int tan^5(x)*sec^4(x)dx$ ?

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Answer 1 · 2016-07-05T17:14:11

$= \frac{1}{8} {tan}^{8} (x) + \frac{1}{6} {tan}^{6} (x) + C$ $= 1/8tan^8(x) + 1/6tan^6(x) + C$

$int dx qquad tan^5(x)*sec^4(x)$

$= int dx qquad tan^5(x)*color{red}{sec^2(x)}*sec^2(x)$

using well known identity....
$= int dx qquad tan^5(x)*color{red}{(tan^2(x) + 1)}*sec^2(x)$

$= int dx qquad (tan^7(x) + tan^5(x))*sec^2(x)$

$= int dx qquad tan^7(x)*sec^2(x) + tan^5(x)*sec^2(x)$

$= 1/8tan^8(x) + 1/6tan^6(x) + C$

What is the integral of int tan^5(x)*sec^4(x)dx∫tan5(x)⋅sec4(x)dxint tan^5(x)*sec^4(x)dx?