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Question #a5f39

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Oct 3, 2017

${sec}^{4} (x) + C$ $sec^4(x)+C$

Explanation:

$\int 4 {sec}^{4} (x) tan (x) d x$ $int4sec^4(x)tan(x)dx$

$= 4 \int {sec}^{3} (x) [sec (x) tan (x) d x]$ $=4intsec^3(x)[sec(x)tan(x)dx]$

Let $u = sec (x)$ $u=sec(x)$ . Differentiating shows that $d u = sec (x) tan (x) d x$ $du=sec(x)tan(x)dx$ , which is already in the integrand.

$= 4 \int u^{3} d u$ $=4intu^3du$

$= 4 (\frac{u^{4}}{4}) + C$ $=4(u^4/4)+C$

$= {sec}^{4} (x) + C$ $=sec^4(x)+C$

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