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Question Number 31073 by abdo imad last updated on 02/Mar/18

$$findI={\int }_0^{\frac{\pi }{2}}\frac{1-sin\theta }{cos\theta }d\theta .$$

Commented by prof Abdo imad last updated on 03/Mar/18

$$thech.tan\left(\frac{\theta }{2}\right)=xgive$$$$I={\int }_0^1\frac{1-\frac{2x}{1+x^2}}{\frac{1-x^2}{1+x^2}}\frac{2dx}{1+x^2}=2{\int }_0^1\frac{{(x-1)}^2}{(1-x^2)(1+x^2)}dx$$$$=2{\int }_0^1\frac{1-x}{(1+x)(1+x^2)}dxletdecomposethefrsction$$$$F\left(x\right)=\frac{1-x}{(1+x)(1+x^2)}=\frac{a}{1+x}+\frac{bx+c}{1+x^2}$$$$a={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{2}{2}=1$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=-a=-1\Rightarrow$$$$F\left(x\right)=\frac{1}{1+x}+\frac{-x+c}{1+x^2}$$$$F\left(0\right)=1=1+c\Rightarrow c=0soF\left(x\right)=\frac{1}{1+x}-\frac{x}{1+x^2}\Rightarrow$$$$I=2{\int }_0^1\frac{dx}{1+x}dx+{\int }_0^1\frac{2x}{1+x^2}dxp$$$$=2{[ln\mid 1+x\mid ]}_0^1-{\left[ln(1+x^2)\right]}_0^1=2ln\left(2\right)-ln\left(2\right)\Rightarrow$$$$I=ln\left(2\right).$$

Answered by Joel578 last updated on 02/Mar/18

$$I=\underset{t\to \frac{\pi }{2}}{\lim }({\int }_0^t\sec x-\tan xdx)$$$$=\underset{t\to \frac{\pi }{2}}{\lim }{\left[\ln (\sec x+\tan x)\cos x\right]}_0^t$$$$=\underset{t\to \frac{\pi }{2}}{\lim }\ln \left((\sec t+\tan t)\cos t\right)-\ln (1+0)$$$$=\underset{t\to \frac{\pi }{2}}{\lim }\ln \left((\sec t+\tan t)\cos t\right)$$$$=\underset{t\to \frac{\pi }{2}}{\lim }\ln (1+\tan t\cos t)=\ln (1+\underset{t\to \frac{\pi }{2}}{\lim }\tan t\cos t)$$$$=\ln (1+\underset{t\to \frac{\pi }{2}}{\lim }\frac{\cos t}{\cot t})$$$$=\ln (1+\underset{t\to \frac{\pi }{2}}{\lim }\frac{-\sin t}{-{\mathrm{cosec}}^{2}t})=\ln 2$$

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