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Question Number 156869 by cortano last updated on 16/Oct/21

Answered by gsk2684 last updated on 16/Oct/21

$$\int \frac{\sec x\left(\sec x(\sec x+\tan x)\right)}{{(\sec x+\tan x)}^6}dx$$$$put\sec x+\tan x=t,\sec x-\tan x=\frac{1}{t}\Rightarrow$$$$\sec x\tan x+\sec {}^{2}x=\frac{dt}{dx},2\sec x=t+\frac{1}{t}$$$$\int \frac{t+\frac{1}{t}}{2t^6}dt=\frac{1}{2}\int (t^{-5}+t^{-7})dt$$$$=\frac{1}{2}(\frac{t^{-4}}{-4}+\frac{t^{-6}}{-6})+const$$$$=-\frac{1}{8{(\sec x+\tan x)}^4}-\frac{1}{12{(\sec x+\tan x)}^6}+const$$

Commented by cortano last updated on 16/Oct/21

$$yes$$

Answered by puissant last updated on 16/Oct/21

$$Q=\int \frac{{sec}^{2}x}{{(secx+tanx)}^5}dx=\int \frac{{sec}^{2}x}{{(\frac{1}{cosx}+\frac{sinx}{cosx})}^5}dx$$$$=\int \frac{{cos}^{3}x}{{(1+sinx)}^5}dx=\int \frac{(1-sinx)(1+sinx)cosx}{{(1+sinx)}^5}dx$$$$=\int \frac{1-sinx}{{(1+sinx)}^4}cosdx;u=sinx\to du=cosxdx$$$$\Rightarrow Q=\int \frac{1-u}{{(1+u)}^4}du=\int \frac{1}{{(1+u)}^4}du-\int \frac{u+1-1}{{(1+u)}^4}du$$$$=2\int \frac{1}{{(1+u)}^4}du-\int \frac{1}{{(1+u)}^3}du$$$$\Rightarrow Q=-\frac{2}{3{(1+u)}^3}+\frac{1}{2{(1+u)}^2}+C$$$$\therefore \because Q=-\frac{2}{3{(1+sinx)}^3}+\frac{1}{2{(1+sinx)}^2}+C..$$$$$$$$.............\mathcal{L}epuissant............$$

Commented by cortano last updated on 16/Oct/21

$$yes$$

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