Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 48175 by Abdo msup. last updated on 20/Nov/18

$$calculate{lim}_{x\to 0}{\int }_x^{x^2}\frac{ln(1+t)}{sin\left(t\right)}dt$$

Commented by maxmathsup by imad last updated on 20/Nov/18

$$\mathrm{\exists }c\in ]x,x^2[/A\left(x\right)={\int }_x^{x^2}\frac{ln(1+t)}{sin\left(t\right)}dt=\frac{1}{sinc}{\int }_x^{x^2}ln(1+t)dtbut$$$${\int }_x^{x^2}ln(1+t)dt{=}_{1+t=u}{\int }_{1+x}^{1+x^2}ln\left(u\right)du={[uln\left(u\right)-u]}_{1+x}^{1+x^2}$$$$=(1+x^2)ln(1+x^2)-(1+x^2)-(1+x)ln(1+x)+1+x$$$$\sim (1+x^2)x^2+x-x^2-x(1+x)=x^2+x^4-2x^2=x^4-x^2$$$$x<c<x^2\Rightarrow \mathrm{\exists }\alpha /c=\alpha x+(1-\alpha )x^{2}andsinc\sim c\Rightarrow$$$$\frac{x^4-x^2}{\alpha x(1-\alpha )x^2}=\frac{x^3-x}{\alpha +(1-\alpha )x}\to 0so{lim}_{x\to 0}A\left(x\right)=0$$

Commented by maxmathsup by imad last updated on 20/Nov/18

$$\alpha \in ]0,1[.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Nov/18

$$I={\int }_x^{x^2}\frac{ln(1+t)}{sint}dt$$$$\frac{dI}{dx}={\int }_x^{x^2}\frac{\partial }{\partial x}\left(\frac{ln(1+t)}{sint}\right)dt+\frac{ln(1+x^2)}{{sinx}^2}\times \frac{{dx}^2}{dx}-\frac{ln(1+x)}{sinx}\times \frac{dx}{dx}$$$$=2x\times \frac{ln(1+x^2)}{{sinx}^2}-\frac{ln(1+x)}{sinx}$$$$=2x\times \frac{ln(1+x^2)}{x^2}\times \frac{1}{\frac{{sinx}^2}{x^2}}-\frac{ln(1+x)}{x}\times \frac{1}{\frac{sinx}{x}}$$$$whenx\to 0$$$$=2\times 0\times \frac{1}{1}-1=-\frac{dI}{dx}$$$$\frac{dI}{dx}=-1I=-x+c$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Nov/18

$$plschecksir...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com