Question #de65e

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Question #de65e

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Answer 1 · 2017-10-01T17:24:57

$\frac{9}{10} sec x \cdot tan x + \frac{9}{10} ln | sec x + tan x | + C$ $9/10 secx*tanx+9/10ln|secx+tanx|+C$

Explanation:

$I = 9 \int sec x \cdot {sec}^{2} x d x$ $I=9intsecx*sec^2xdx$
Using integration by parts
$= 9 sec x \int {sec}^{2} x d x - 9 \int (\frac{d}{d x} sec x \cdot \int {sec}^{2} x d x) d x$ $=9secx intsec^2xdx-9int(d/dx secx*intsec^2xdx)dx$
$= 9 sec x \cdot tan x - 9 \int (sec x \cdot tan x \cdot tan x) d x$ $=9secx*tanx-9int(secx*tanx*tanx)dx$
$= 9 sec x \cdot tan x - 9 \int (sec x \cdot {tan}^{2} x) d x$ $=9secx*tanx-9int(secx*tan^2x)dx$
$= 9 sec x \cdot tan x - 9 \int (sec x \cdot ({sec}^{2} x - 1) d x$ $=9secx*tanx-9int(secx*(sec^2x-1)dx$
$I = 9 sec x \cdot tan x - 9 \int {sec}^{3} x d x + 9 \int sec x d x$ $I=9secx*tanx-9intsec^3xdx+9intsecxdx$
$I = 9 sec x \cdot tan x - 9 I + 9 \int sec x d x$ $I=9secx*tanx-9I+9intsecxdx$
$10 I = 9 sec x \cdot tan x + 9 ln | sec x + tan x | + C$ $10I=9secx*tanx+9ln|secx+tanx|+C$
$I = \frac{9}{10} sec x \cdot tan x + \frac{9}{10} ln | sec x + tan x | + C$ $I=9/10 secx*tanx+9/10 ln|secx+tanx|+C$

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Question #de65e

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