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Question Number 83331 by oyemi kemewari last updated on 01/Mar/20

Commented by jagoll last updated on 01/Mar/20

$$\left(13\right)\int \sec \mathrm{x}\mathrm{dx}=\int \frac{\sec \mathrm{x}(\sec \mathrm{x}+\tan \mathrm{x})\mathrm{dx}}{\sec \mathrm{x}+\tan \mathrm{x}}$$$$=\int \frac{\sec {}^2\mathrm{x}+\sec \mathrm{x}\tan \mathrm{x}\mathrm{dx}}{\sec \mathrm{x}+\tan \mathrm{x}}$$$$=\int \frac{\mathrm{d}(\sec \mathrm{x}+\tan \mathrm{x})}{\sec \mathrm{x}+\tan \mathrm{x}}$$$$=\ln \mid \sec \mathrm{x}+\tan \mathrm{x}\mid +\mathrm{c}$$

Commented by mr W last updated on 01/Mar/20

$$\left(1\right):$$$$maximum=\sqrt{\underset{i=1}{\sum ^n}{a}_i^2}$$$$seeQ83341$$

Commented by abdomathmax last updated on 01/Mar/20

$$15)weusethediffeomorphism(r,\theta )\to (x,y)/$$$$x=rcos\theta andy=rsin\theta wehavex,y⩾0\Rightarrow 0⩽\theta ⩽\frac{\pi }{2}$$$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(x^2+y^2)}dxdy={\int }_0^{\mathrm{\infty }}{\int }_0^{\frac{\pi }{2}}{e}^{-r^2}rdrd\theta$$$$=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}r{e}^{-r^2}dr=\frac{\pi }{2}{[-\frac{1}{2}{e}^{-r^2}]}_0^{+\mathrm{\infty }}=\frac{\pi }{4}$$

Commented by abdomathmax last updated on 01/Mar/20

$$18)A={\int }_0^a{\int }_0^a\frac{xdxdy}{\sqrt{x^2+y^2}}(wesupposea>0)weuse$$$$thediffeomorphism(r,\theta )\to (x,y)=(rcos\theta ,rsin\theta )\Rightarrow$$$$0⩽x,y⩽a\Rightarrow 0⩽x^2,y^2⩽a^2\Rightarrow 0⩽x^2+y^2⩽2a^2\Rightarrow$$$$0⩽r^2⩽2a^2\Rightarrow 0⩽r⩽a\sqrt{2}\Rightarrow$$$$A={\int }_0^{a\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\frac{rcos\theta rdrd\theta }{r}$$$$={\int }_0^{a\sqrt{2}}rdr{\int }_0^{\frac{\pi }{2}}d\theta =\frac{\pi }{2}{\left[\frac{r^2}{2}\right]}_0^{a\sqrt{2}}=\frac{\pi }{2}\left(\frac{2a^2}{2}\right)=\frac{\pi a^2}{2}$$

Commented by mathmax by abdo last updated on 01/Mar/20

$$\int 5^{lnx}dx=Ichangementlnx=tgive$$$$I=\int 5^t{e}^{t}dt=\int e^t\times e^{tln\left(5\right)}dt=\int e^{(1+ln5)t}dt$$$$=\frac{1}{1+ln5}{e}^{(1+ln5)t}+c=\frac{1}{1+ln5}{e}^{(1+ln5)lnx}=\frac{1}{1+ln5}.x.5^x+c$$

Commented by mathmax by abdo last updated on 02/Mar/20

$$23)A_n={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-xy}sin\left(nx\right)dxdy={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{e}^{-yx}sin\left(nx\right)dx\right)dy$$$${\int }_0^{\mathrm{\infty }}{e}^{-yx}sin\left(nx\right)=Im\left({\int }_0^{\mathrm{\infty }}{e}^{-yx+inx}dx\right)$$$${\int }_0^{\mathrm{\infty }}{e}^{(-y+in)x}dx={\left[\frac{1}{-y+in}{e}^{(-y+in)x}\right]}_0^{+\mathrm{\infty }}=-\frac{1}{-y+in}=\frac{1}{y-in}$$$$=\frac{y+in}{y^2+n^2}\Rightarrow {\int }_0^{\mathrm{\infty }}{e}^{-yx}sin\left(nx\right)dx=\frac{n}{y^2+n^2}\Rightarrow$$$$A_n=n{\int }_0^{\mathrm{\infty }}\frac{dy}{y^2+n^2}{=}_{y=nu}n{\int }_0^{\mathrm{\infty }}\frac{ndu}{n^2(1+u^2)}=\frac{\pi }{2}byfubinniwehave$$$$A_n={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{e}^{-xy}dy\right)sin\left(nx\right)dx$$$$={\int }_0^{\mathrm{\infty }}{[-\frac{1}{x}{e}^{-xy}]}_{y=0}^{\mathrm{\infty }}sin\left(nx\right)dx={\int }_0^{\mathrm{\infty }}\frac{sin\left(nx\right)}{x}dx=\frac{\pi }{2}$$$$$$

Commented by mathmax by abdo last updated on 02/Mar/20

$$I={\int }_0^{\mathrm{\infty }}\frac{1-e^{-ax}}{x}{e}^{-x}dx\Rightarrow I={\int }_0^{\mathrm{\infty }}\frac{e^{-x}-e^{-(a+1)x}}{x}dx=\phi \left(a\right)$$$${\phi }^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-(a+1)x}dx={[-\frac{1}{a+1}{e}^{-(a+1)x}]}_0^{+\mathrm{\infty }}=\frac{1}{a+1}\Rightarrow$$$$\phi \left(a\right)=k+ln(a+1)wehave\phi \left(0\right)=0=k\Rightarrow$$$$I=ln(a+1)\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{1-e^{-x}}{x}{e}^{-x}dx={\int }_0^{\mathrm{\infty }}\frac{e^{-x}-e^{-2x}}{x}dx=ln\left(2\right)$$

Answered by mind is power last updated on 02/Mar/20

$$13){\int }_0^{+\mathrm{\infty }}\frac{1-e^{-ax}}{x}{e}^{-x}dx$$$$={\int }_0^{+\mathrm{\infty }}\frac{e^{-x}-e^{-(a+1)x}}{x}dx=ln\left(\frac{a+1}{1}\right)=ln(a+1)$$$${\int }_0^{+\mathrm{\infty }}\frac{-\underset{k=1}{\sum ^{+\mathrm{\infty }}}\frac{{(-ax)}^k{e}^{-x}}{k!}}{x}$$$$=\sum _{k⩾1}{(-1)}^{k+1}.\frac{a^k}{k!}.{\int }_0^{+\mathrm{\infty }}{x}^{k-1}.e^{-x}dx$$$$=\sum _{k⩾1}\frac{{(-1)}^{k+1}}{k!}{a}^k.\mathrm{\Gamma }\left(k\right)=\sum _{k⩾1}\frac{{(-1)}^{k+1}}{k}{a}^k=ln(1+a),\mathrm{\forall }a\in ]0,1[$$

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