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Question Number 212239 by Spillover last updated on 07/Oct/24

Answered by Ghisom last updated on 08/Oct/24

$$\int \frac{{\mathrm{e}}^{-x/2}\sqrt{1-\sin x}}{1+\cos x}dx=$$$$[t=\tan \frac{x}{2}]$$$$=\int \frac{{\mathrm{e}}^{-\arctan t}(t-1)}{\sqrt{t^2+1}}dt=$$$$=\int \frac{{\mathrm{e}}^{-\arctan t}t}{\sqrt{t^2+1}}dt-\int \frac{{\mathrm{e}}^{-\arctan t}}{\sqrt{t^2+1}}dt=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$={\mathrm{e}}^{-\arctan t}\sqrt{t^2+1}+\int \frac{{\mathrm{e}}^{-\arctan t}}{\sqrt{t^2+1}}dt-\int \frac{{\mathrm{e}}^{-\arctan t}}{\sqrt{t^2+1}}dt=$$$$={\mathrm{e}}^{-\arctan t}\sqrt{t^2+1}=$$$$=\frac{{\mathrm{e}}^{-x/2}}{\cos \frac{x}{2}}+C$$

Commented by Frix last updated on 08/Oct/24

$$\mathrm{Nice}!$$

Commented by Spillover last updated on 08/Oct/24

$$great.thanks$$

Commented by MathematicalUser2357 last updated on 12/Oct/24

$$\mathrm{No},\mathrm{the}\mathrm{real}\mathrm{answer}\mathrm{is}\frac{e^{-x/2}\sqrt{1-\sin x}{\sec }^2\frac{x}{2}}{\tan \frac{x}{2}+1}+C.$$

Commented by Ghisom last updated on 13/Oct/24

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same},\mathrm{just}\mathrm{transform}\mathrm{it}\mathrm{to}\mathrm{see}$$

Answered by MathematicalUser2357 last updated on 12/Oct/24

$$\frac{e^{-x/2}\sqrt{1-\sin x}{\sec }^2\frac{x}{2}}{\tan \frac{x}{2}-1}+C$$

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