Question Number 187662 by Jonas007 last updated on 20/Feb/23
Answered by Ar Brandon last updated on 20/Feb/23
$${I}=\int\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}{x}+\mathrm{cos}{x}\right)^{\mathrm{2}} }{dx}=\int\frac{{x}\mathrm{cos}{x}}{\left({x}\mathrm{sin}{x}+\mathrm{cos}{x}\right)^{\mathrm{2}} }\centerdot\frac{{x}}{\mathrm{cos}{x}}{dx} \\ $$$$\begin{cases}{\mathrm{u}\left({x}\right)=\frac{{x}}{\mathrm{cos}{x}}}\\{\mathrm{v}'\left({x}\right)=\frac{{x}\mathrm{cos}{x}}{\left({x}\mathrm{sin}{x}+\mathrm{cos}{x}\right)^{\mathrm{2}} }}\end{cases}\:\Rightarrow\begin{cases}{\mathrm{u}'\left({x}\right)=\frac{\mathrm{cos}{x}+{x}\mathrm{sin}{x}}{\mathrm{cos}^{\mathrm{2}} {x}}}\\{\mathrm{v}\left({x}\right)=−\frac{\mathrm{1}}{{x}\mathrm{sin}{x}+\mathrm{cos}{x}}}\end{cases} \\ $$$${I}=−\frac{\frac{{x}}{\mathrm{cos}{x}}}{{x}\mathrm{sin}{x}+\mathrm{cos}{x}}+\int\frac{{dx}}{\mathrm{cos}^{\mathrm{2}} {x}}=\mathrm{tan}{x}−\frac{\frac{{x}}{\mathrm{cos}{x}}}{{x}\mathrm{sin}{x}+\mathrm{cos}{x}}+{C} \\ $$
Commented by CElcedricjunior last updated on 20/Feb/23
$$\bigstar \\ $$
Commented by Ar Brandon last updated on 20/Feb/23
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