How do you integrate ${sec}^{2} x {tan}^{10} x$ $sec^2xtan^10x$ ?

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Answer 1 · 2016-06-20T06:54:58

$\int {sec}^{2} x {tan}^{10} x d x = \frac{{tan}^{11} x}{11} + c$ $intsec^2xtan^(10)xdx=tan^11x/11+c$

Let $u = tan x$ $u=tanx$ , then $d u = {sec}^{2} x d x$ $du=sec^2xdx$

Hence, $\int {sec}^{2} x {tan}^{10} x d x = \int u^{10} d u$ $intsec^2xtan^(10)xdx=intu^10du$

= $\frac{u^{11}}{11} + c = \frac{{tan}^{11} x}{11} + c$ $u^11/11+c=tan^11x/11+c$

How do you integrate sec2xtan10xsec^2xtan^10x?