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Question Number 40142 by maxmathsup by imad last updated on 16/Jul/18

$$calculate{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{1+tan\left(x\right)}{1+sin\left(2x\right)}dx$$$$$$

Commented by maxmathsup by imad last updated on 19/Jul/18

$$letI={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{1+tan\left(x\right)}{1+sin\left(2x\right)}dxchangementtanx=tgive$$$$I={\int }_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{1+t}{1+\frac{2t}{1+t^2}}\frac{dt}{1+t^2}={\int }_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{1+t}{1+t^2+2t}dt$$$$={\int }_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{t+1}{{(t+1)}^2}dt={\int }_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{dt}{t+1}={[ln\mid t+1\mid ]}_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}$$$$=ln(1+\frac{1}{\sqrt{3}})-ln(1-\frac{1}{\sqrt{3}})=ln(\sqrt{3}+1)-ln\left(\sqrt{3}\right)-ln(\sqrt{3}-1)+ln\left(\sqrt{3}\right)$$$$I=ln(\sqrt{3}+1)-ln(\sqrt{3}-1)$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18

$$t=tanxdt={sec}^{2}xdx$$$${\int }_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{(1+t)(1+t^2)(1+t^2)dt}{1+t^2+2t}$$$${\int }_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{t^4+2t^2+1}{t+1}dt$$$${\int }_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{t^4+t^3-t^3-t^2+3t^2+3t-3t-3+4}{t+1}$$$${\int }_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}{t}^3-t^2+3t-3+\frac{4}{t+1}dt$$$$=\mid \frac{t^4}{4}-\frac{t^3}{3}+3\frac{t^2}{2}-3t+4ln(t+1){\mid }_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}$$$$=-\frac{1}{3}\left(\frac{2}{3\sqrt{3}}\right)-3\left(\frac{2}{\sqrt{3}}\right)+4ln\left(\frac{\frac{1}{\sqrt{3}}+1}{\frac{-1}{\sqrt{3}}+1}\right)$$$$=\frac{-2}{9\sqrt{3}}-\frac{6}{\sqrt{3}}+4ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$$$$$$$$$$$$0$$$$$$

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