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Question Number 169164 by 0731619 last updated on 25/Apr/22

Commented by infinityaction last updated on 25/Apr/22

$$y=\ln (\sqrt{\frac{(1-\sin x)/\cos x}{(1+\sin x)/\cos x}}$$$$y=\ln \sqrt{\frac{(\sec x-\tan x)(\sec x-\tan x)}{(\sec x+\tan x)(\sec x-\tan x)}}$$$$y=\ln (\sec x-\tan x)$$$$\frac{dy}{dx}=\frac{\sec x(\tan x-\sec x)}{(\sec x-\tan x)}$$$$\frac{dy}{dx}=-\sec x$$$$$$

Commented by infinityaction last updated on 25/Apr/22

$$let$$$$p=\frac{x^2+3x}{(x+1)(x^2+1)}$$$$p=\frac{{(x+1)}^2+(x-1)}{(x+1)(x^2+1)}$$$$p=\frac{x+1}{x^2+1}+\frac{x-1}{(x+1)(x^2+1)}$$$$p=\frac{x+1}{x^2+1}-\frac{(x^2+1)-x(x+1)}{(x+1)(x^2+1)}$$$$p=\frac{x+1}{x^2+1}-\frac{1}{x+1}+\frac{x}{x^2+1}$$$$p=\frac{2x}{x^2+1}+\frac{1}{x^2+1}-\frac{1}{x+1}$$$$I={\int }_0^1\{\frac{2x}{x^2+1}+\frac{1}{x^2+1}-\frac{1}{x+1}\}dx$$$$I={[\log (x^2+1)+{\tan }^{-1}\left(x\right)+\log (x+1)]}_0^1$$$$I=\log 2+\frac{\pi }{4}+\log 2$$$$I=\frac{\pi }{4}+\log 4$$$$$$$$$$

Answered by Mathspace last updated on 25/Apr/22

$$$$$$1)F\left(x\right)=ln\left(\sqrt{\frac{1-cos(\frac{\pi }{2}-x)}{1+cos(\frac{\pi }{2}-x)}}\right)$$$$=ln\sqrt{\frac{2{sin}^2(\frac{\pi }{4}-\frac{x}{2})}{2{cos}^2(\frac{\pi }{4}-\frac{x}{2})}}$$$$=ln\mid tan(\frac{\pi }{4}-\frac{x}{2})\Rightarrow$$$$F^{{}^{\prime }}\left(x\right)=\frac{-\frac{1}{2}(1+{tan}^2(\frac{\pi }{4}-\frac{x}{2}))}{tan(\frac{\pi }{4}-\frac{x}{2})}$$

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