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Question Number 214853 by liuxinnan last updated on 21/Dec/24

$$\int \frac{1+\cos x}{1+\sin {}^{2}x}dx=?$$

Commented by liuxinnan last updated on 21/Dec/24

$$Ihadlongtimenotlookthrough$$$$thisapp,Ijustfind$$

Answered by TonyCWX08 last updated on 21/Dec/24

Commented by TonyCWX08 last updated on 21/Dec/24

$$If{it}^{\prime }stooblurry,pleasemention.$$$$I^{\prime }llwritetheminLaTeXform.$$

Commented by liuxinnan last updated on 25/Dec/24

$$thanks$$

Answered by Frix last updated on 21/Dec/24

$$\int \frac{1+\cos x}{1+{\sin }^{2}x}dx=\int \frac{1+\cos x}{2-{\cos }^{2}x}dx=$$$$=\frac{2+\sqrt{2}}{4}\int \frac{dx}{\sqrt{2}-\cos x}-\frac{2-\sqrt{2}}{4}\int \frac{dx}{\sqrt{2}+\cos x}=$$$$[t=\tan \frac{x}{2}]$$$$=\frac{2+\sqrt{2}}{2}\int \frac{dt}{(1+\sqrt{2})t^2-1+\sqrt{2}}+\frac{2-\sqrt{2}}{2}\int \frac{dt}{(1-\sqrt{2})t^2-1-\sqrt{2}}=$$$$=\frac{2+\sqrt{2}}{2}{\tan }^{-1}\left((1+\sqrt{2})t\right)-\frac{2-\sqrt{2}}{2}{\tan }^{-1}\left((1-\sqrt{2})t\right)=$$$$==\frac{2+\sqrt{2}}{2}{\tan }^{-1}\left((1+\sqrt{2})\tan \frac{x}{2}\right)-\frac{2-\sqrt{2}}{2}{\tan }^{-1}\left((1-\sqrt{2})\tan \frac{x}{2}\right)+C$$

Commented by liuxinnan last updated on 25/Dec/24

$$thanks$$

Answered by MrGaster last updated on 22/Dec/24

$$\int \frac{1}{1+{\sin }^{2}x}dx+\int \frac{\cos x}{1+{\sin }^{2}x}dx$$$$\int \frac{1}{{\sin }^{2}x}dx\Rrightarrow {\sin }^{2}x=1-\cos x$$$$\therefore \Rrightarrow \int \frac{1}{2-{\cos }^{2}x}dx+\int \frac{\cos x}{1+{\sin }^{2}x}dx\Rrightarrow 2-{\cos }^{2}x=2-\frac{1}{{\sec }^{2}x}=\frac{{2\sec }^{2}x-1}{{\sec }^{2}x}$$$$\therefore \Rrightarrow \int \frac{{\sec }^{2}x}{{2\sec }^{2}x-1}dx+\int \frac{\cos x}{1+{\sin }^{2}x}dx\Rrightarrow \int \frac{{\sec }^{2}x}{1+{\left(\sqrt{2}\tan x\right)}^2}dx+\int \frac{d\left(\sin x\right)}{1+{\sin }^{2}x}$$$$=\frac{1}{\sqrt{2}}\arctan \left(\sqrt{2}\tan x\right)+\arctan \left(\sin x\right)+C$$

Commented by liuxinnan last updated on 25/Dec/24

$$thanks$$

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