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Question Number 166449 by cortano1 last updated on 20/Feb/22

$$\mathrm{C}=\int \frac{1-\tan {}^2\mathrm{x}}{1+\sec {}^2\mathrm{x}}\mathrm{dx}=?$$

Commented by cortano1 last updated on 20/Feb/22

$$\mathrm{oo}\mathrm{yes}$$

Commented by cortano1 last updated on 20/Feb/22

Commented by MJS_new last updated on 20/Feb/22

$$u=\tan \frac{x}{2}\iff x=2\arctan u\Rightarrow {\cos }^{2}x={\left(\frac{1-u^2}{1+u^2}\right)}^2$$

Answered by MJS_new last updated on 20/Feb/22

$$\int \frac{1-{\tan }^{2}x}{1+{\sec }^{2}x}dx=2\int dx-3\int \frac{1}{1+{\cos }^{2}x}dx=$$$$[t=\tan x\to dx={\cos }^{2}xdt]$$$$=2x-3\int \frac{dt}{t^2+2}=$$$$=2x-\frac{3}{\sqrt{2}}\arctan \frac{t}{\sqrt{2}}=$$$$=2x-\frac{3}{\sqrt{2}}\arctan \frac{\tan x}{\sqrt{2}}+C$$

Commented by peter frank last updated on 24/Feb/22

$$\mathrm{great}$$

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