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Question Number 212099 by vahid last updated on 30/Sep/24

Answered by mehdee7396 last updated on 30/Sep/24

$$\int \frac{1+{tan}^2\frac{x}{2}}{1-{tan}^2\frac{x}{2}}dx;lettan\frac{x}{2}=u$$$$=\int \frac{2u}{1-u^2}du=ln\frac{1+u}{1-u}+c$$$$=ln\left(tan(\frac{\pi }{4}+\frac{x}{2})\right)+c✓$$$$$$$$$$

Answered by Frix last updated on 30/Sep/24

$$\int \frac{dx}{\cos x}=\int \sec xdx=$$$$=\int \sec x\frac{\tan x+\sec x}{\tan x+\sec x}dx\stackrel{[t=\tan x+\sec x]}{=}$$$$=\int \frac{dt}{t}=\ln t=$$$$=\ln \mid \tan x+\sec x\mid +C$$

Answered by mr W last updated on 30/Sep/24

$$=\int \frac{d\left(\sin x\right)}{1-{\sin }^{2}x}$$$$=\frac{1}{2}[\int \frac{d\left(\sin x\right)}{1-\sin x}+\int \frac{d\left(\sin x\right)}{1+\sin x}]$$$$=\frac{1}{2}\ln \frac{1+\sin x}{1-\sin x}+C$$

Commented by MathematicalUser2357 last updated on 02/Oct/24

1/2 ln|(1+sinx)/(1-sinx)|+C*

Commented by Frix last updated on 02/Oct/24

$$\mathrm{What}?\mathrm{Why}?$$$$-1⩽\sin x⩽1$$$$1.0⩽1+\sin x⩽2$$$$2.-2⩽-1+\sin x⩽0\iff 0⩽1-\sin x⩽2$$$$\Rightarrow$$$$\mid \frac{1+\sin x}{1-\sin x}\mid =\frac{1+\sin x}{1-\sin x}$$

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