tích phân

N

nguyenbahiep1

$\begin{cases} u = x \\ dv = \frac{sinxdx}{(1+cosx)^2} \end{cases} \Rightarrow \begin{cases} du = dx \\ v = \frac{1}{1+cosx} \end{cases} \\ I = \frac{x}{1+cosx} \big|_0^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+cosx} \\ I = \frac{\pi}{2} - \int_{0}^{\frac{\pi}{2}} \frac{dx}{2cos^2(\frac{x}{2})} \\ \Rightarrow I = \frac{\pi}{2} - tan(\frac{x}{2}) \big|_0^{\frac{\pi}{2}} = ?$
 
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