Question #9e6e3

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Question #9e6e3

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Answer 1 · 2018-02-20T09:37:50

$I = \frac{1}{5} {tan}^{5} (x) + \frac{2}{3} {tan (x)}^{3} + tan (x) + C$ $I=1/5tan^5(x)+2/3tan(x)^3+tan(x)+C$

Explanation:

We want to solve

$I = \int \frac{1}{{cos}^{6} (x)} d x = \int {sec}^{2} (x) {({sec}^{2} (x))}^{2} d x$ $I=int1/cos^6(x)dx=intsec^2(x)(sec^2(x))^2dx$

Use the identity ${sec}^{2} (x) = 1 + {tan}^{2} (x)$ $sec^2(x)=1+tan^2(x)$

$I = \int {sec}^{2} (x) {({tan}^{2} (x) + 1)}^{2} d x$ $I=intsec^2(x)(tan^2(x)+1)^2dx$

Make a substitution $u = tan (x) \Rightarrow \frac{d u}{d x} = {sec}^{2} (x)$ $u=tan(x)=>(du)/dx=sec^2(x)$

$I = \int {(u^{2} + 1)}^{2} d u$ $I=int(u^2+1)^2du$

$= \int u^{4} + 2 u^{2} + 1 d u$ $=intu^4+2u^2+1du$

$= \frac{1}{5} u^{5} + \frac{2}{3} u^{3} + u + C$ $=1/5u^5+2/3u^3+u+C$

Substitute back $u = tan (x)$ $u=tan(x)$

$I = \frac{1}{5} {tan}^{5} (x) + \frac{2}{3} {tan (x)}^{3} + tan (x) + C$ $I=1/5tan^5(x)+2/3tan(x)^3+tan(x)+C$

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Question #9e6e3

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