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Question Number 62814 by Cheyboy last updated on 25/Jun/19

Commented by Cheyboy last updated on 25/Jun/19

$$Pleasehelpmewiththesequestion$$

Commented by mathmax by abdo last updated on 25/Jun/19

$$3)letI={\int }_0^{\mathrm{\infty }}\frac{16arctan\left(x\right)}{1+x^2}dx\Rightarrow I=16{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{1+x^2}dxbyparts$$$${\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{1+x^2}dx={\left[{arctan}^{2}x\right]}_0^{+\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{1+x^2}dx=\frac{{\pi }^2}{4}-{\int }_0^{\mathrm{\infty }}\frac{arctanx}{1+x^2}dx$$$$2{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{1+x^2}dx=\frac{{\pi }^2}{4}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{1+x^2}dx=\frac{{\pi }^2}{8}\Rightarrow I=16.\frac{{\pi }^2}{8}\Rightarrow I=2{\pi }^2.$$

Commented by Prithwish sen last updated on 25/Jun/19

$$\mathrm{a})\int \frac{\mathrm{du}}{{\mathrm{u}}^2+3\mathrm{u}+2}\mathrm{putting}{\mathrm{e}}^{\mathrm{t}}=\mathrm{u}\Rightarrow {\mathrm{e}}^{\mathrm{t}}\mathrm{dt}=\mathrm{du}$$$$\int \frac{\mathrm{du}}{(\mathrm{u}+2)(\mathrm{u}+1)}=\int \frac{1}{\mathrm{u}+1}-\frac{1}{\mathrm{u}+2}\mathrm{du}$$$$=\ln \frac{{\mathrm{e}}^{\mathrm{t}}+1}{{\mathrm{e}}^{\mathrm{t}}+2}+\mathrm{C}$$$$$$

Commented by mathmax by abdo last updated on 25/Jun/19

$$2b)letA=\int \frac{e^t}{\sqrt{e^{2t}+9}}dtchangememet{e}^t=x\Rightarrow t=ln\left(x\right)$$$$A=\int \frac{x}{\sqrt{x^2+9}}\frac{dx}{x}=\int \frac{dx}{\sqrt{x^2+9}}{=}_{x=3u}\int \frac{3du}{3\sqrt{1+u^2}}=ln(u+\sqrt{1+u^2}+c$$$$=ln(\frac{x}{3}+\sqrt{1+\frac{x^2}{9}})+c=ln\left(\frac{x+\sqrt{x^2+9}}{3}\right)+c$$$$=ln(x+\sqrt{x^2+9})+C=ln(e^t+\sqrt{e^{2t}+9})+C.$$

Commented by Cheyboy last updated on 25/Jun/19

$$\mathbb{THANK}\mathbb{YOU}\mathbb{ALL}\mathbb{SIRS}$$$$\mathbb{FOR}\mathbb{HELP}\mathbb{WELL}\mathbb{APPRECIATED}$$

Commented by Prithwish sen last updated on 25/Jun/19

$$2\mathrm{b})\int {\left(2\mathrm{cosysiny}\right)}^2{2\sin }^2\mathrm{ydy}$$$$=\int {\sin }^{2}2\mathrm{y}(1-\cos 2\mathrm{y})\mathrm{dy}$$$$=\int \frac{1}{2}(1-\cos 4\mathrm{y})\mathrm{dy}-\int {\sin }^{2}2\mathrm{ycos}2\mathrm{y}\mathrm{dy}$$$$={\mathrm{I}}_1+{\mathrm{I}}_2$$$${\mathrm{I}}_1=\frac{\mathrm{y}}{2}-\frac{\sin 4\mathrm{y}}{8}$$$${\mathrm{I}}_2=\int {\sin }^{2}2\mathrm{ycos}2\mathrm{y}\mathrm{dy}$$$$\mathrm{put}\sin 2\mathrm{y}=\mathrm{u}\Rightarrow \cos 2\mathrm{y}\mathrm{dy}=\frac{\mathrm{du}}{2}$$$$\int {\mathrm{u}}^2\frac{\mathrm{dy}}{2}=\frac{{\mathrm{u}}^3}{6}=\frac{1}{6}{\sin }^{3}2\mathrm{y}$$$$\therefore \frac{\mathrm{y}}{2}-\frac{\sin 4\mathrm{y}}{8}+\frac{{\sin }^{3}2\mathrm{y}}{6}+\mathrm{C}$$

Commented by Cheyboy last updated on 25/Jun/19

$$Thankalotsir$$

Commented by Cheyboy last updated on 25/Jun/19

Commented by mathmax by abdo last updated on 26/Jun/19

$$1a)letA=\int \frac{e^t}{e^{2t}+3e^t+2}dtchangement{e}^t=xgivet=ln\left(x\right)\Rightarrow$$$$A=\int \frac{x}{x^2+3x+2}\frac{dx}{x}=\int \frac{dx}{x^2+3x+2}$$$$x^2+3x+2=0\to \mathrm{\Delta }=9-8=1\Rightarrow x_1=\frac{-3+1}{2}=-1and{x}_2=\frac{-3-1}{2}=-2\Rightarrow$$$$A=\int \frac{dx}{(x+1)(x+2)}=\int (\frac{1}{x+1}-\frac{1}{x+2})d=ln\mid x+1\mid -ln\mid x+2\mid +c$$$$=ln\mid e^t+1\mid -ln\mid e^t+2\mid +c=ln(1+e^t)-ln(2+e^t)+c.$$

Commented by mathmax by abdo last updated on 27/Jun/19

$$b)letI=\int \frac{cosx}{{sin}^{2}x+sinx-6}dxchangementsinx=tgive$$$$I=\int \frac{dt}{t^2+t-6}letslve{t}^2+t-6\to \mathrm{\Delta }=1-4(-6)=25\Rightarrow t_1=\frac{-1+5}{2}=2$$$$and{t}_2=\frac{-1-5}{2}=-3\Rightarrow I=\int \frac{dt}{(t-2)(t+3)}=\frac{1}{5}\int (\frac{1}{t-2}-\frac{1}{t+3})dt$$$$=\frac{1}{5}\{ln\mid t-2\mid -ln\mid t+3\mid \}+c$$$$=\frac{1}{5}ln\mid \frac{t-2}{t+3}\mid +c=\frac{1}{5}ln\mid \frac{sinx-2}{sinx+3}\mid +c.$$

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