Category: Integration

Question Number 25887 by shivram198922@gmail.com last updated on 16/Dec/17 Answered by ajfour last updated on 16/Dec/17 $${\int }_0^{\pi }[\cos (p-n)x-\cos (p+n)x]dx$$$$=\frac{\sin (p-n)\pi }{p-n}-\frac{\sin (p+n)\pi }{p+n}$$$$=\frac{{p}\left[\mathrm{sin}\:\left({p}−{n}\right)\pi−\mathrm{sin}\:\left({p}+{n}\right)\pi\right]}{{p}^{\mathrm{2}} −{n}^{\mathrm{2}} }…

Question Number 25865 by anonymous_ last updated on 15/Dec/17 $$\int \frac{2+2x}{(x-1)(x^2+1)}dx$$ Answered by ajfour last updated on 16/Dec/17 $$\frac{2+2x}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$$$$\Rightarrow\mathrm{2}+\mathrm{2}{x}={A}\left({x}^{\mathrm{2}}…

Question Number 156914 by cortano last updated on 17/Oct/21 Answered by puissant last updated on 17/Oct/21 $$D=\int \frac{arcsin\left(\frac{1}{x}\right)}{x^5}dx;u=\frac{1}{x}\to du=-\frac{1}{x^2}dx$$$$\Rightarrow\:{D}=−\int\:\frac{{u}^{\mathrm{5}} {arcsin}\left({u}\right)}{{u}^{\mathrm{2}} }{du}=−\int{u}^{\mathrm{3}} {arcsin}\left({u}\right){du} \…

Question Number 25837 by mubeen897@hotmail.com last updated on 15/Dec/17 $$\int \left(x^{n}lnx\right)dx$$ Answered by kaivan.ahmadi last updated on 15/Dec/17 $$\mathrm{u}=\mathrm{lnx}\Rightarrow \mathrm{du}=\frac{\mathrm{dx}}{\mathrm{x}}$$$$\mathrm{dv}=\mathrm{x}^{\mathrm{n}} \mathrm{dx}\Rightarrow\mathrm{v}=\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}…

Question Number 25821 by abdo imad last updated on 15/Dec/17 $${answer}\:{to}\:{q}\mathrm{25796}\:{f}_{} {ind}\:{the}\:{value}\:{off}\left({x}\right)=\:\int_{\mathrm{0}^{} } ^{\pi_{} } {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:{with}\:\mathrm{0}<{x}<\mathrm{1}\:\:\:\partial{f}/\partial{x}=\:\int_{\mathrm{0}} ^{\pi} \:\:{cos}\theta\left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta=\pi{x}^{−\mathrm{1}} −{x}^{−\mathrm{1}} \int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta{and}\:{by}\:{the}\:{changeent}\:{tan}\theta={u}\:{then}\:{the}\:{changement}\:{u}=\left(\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}\right)^{−\mathrm{1}^{} }…