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Question Number 118340 by bramlexs22 last updated on 17/Oct/20

$$\underset{\pi /3}{\stackrel{\pi /2}{\int }}\frac{dx}{1+\sin x-\cos x}$$

Commented by Dwaipayan Shikari last updated on 17/Oct/20

$${\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{dx}{1-cosx+sinx}=2{\int }_{\frac{1}{\sqrt{3}}}^1\frac{dt}{1-\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}.\frac{1}{1+t^2}x=tan\frac{x}{2}$$$$=2{\int }_{\frac{1}{\sqrt{3}}}^1\frac{dt}{2t^2+2t}={\int }_{\frac{1}{\sqrt{3}}}^1\frac{dt}{t(t+1)}={\left[log\left(\frac{t}{t+1}\right)\right]}_{\frac{1}{\sqrt{3}}}^1$$$$=log\left(\frac{1}{2}\right)-log\left(\frac{1}{\sqrt{3}+1}\right)=log\left(\frac{\sqrt{3}+1}{2}\right)$$$$$$

Answered by Bird last updated on 17/Oct/20

$$I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{dx}{1+sinx-cosx}$$$$wedothech.tan\left(\frac{x}{2}\right)=t\Rightarrow$$$$I={\int }_{\frac{1}{\sqrt{3}}}^1\frac{2dt}{(1+t^2)(1+\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2})}$$$$={\int }_{\frac{1}{\sqrt{3}}}^1\frac{2dt}{1+t^2+2t-1+t^2}$$$$={\int }_{\frac{1}{\sqrt{3}}}^1\frac{2dt}{2t^2+2t}={\int }_{\frac{1}{\sqrt{3}}}^1\frac{dt}{t(t+1)}$$$$={\int }_{\frac{1}{\sqrt{3}}}^1(\frac{1}{t}-\frac{1}{t+1})dt$$$$={[ln\mid \frac{t}{t+1}\mid ]}_{\frac{1}{\sqrt{3}}}^1=ln\left(\frac{1}{2}\right)-ln\left(\frac{1}{\sqrt{3}(1+\frac{1}{\sqrt{3}})}\right)$$$$=-ln\left(2\right)+ln(1+\sqrt{3})$$$$$$

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