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Question Number 68434 by mhmd last updated on 10/Sep/19

Answered by mind is power last updated on 10/Sep/19

$${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+\mathrm{b}-\mathrm{x})\mathrm{dx}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{3\sin \left(\mathrm{x}\right)-{2\sin }^2\left(\mathrm{x}\right)}{\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)}$$$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}f\left(x\right)dx={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}f(\frac{\pi }{2}-x)dx={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{3cos\left(x\right)-2{cos}^3\left(x\right)}{sin\left(x\right)+cos\left(x\right)}dx$$$$2{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}f\left(x\right)dx={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{3sin\left(x\right)-2{sin}^2\left(x\right)}{cos\left(x\right)+sin\left(x\right)}dx+{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{3cos\left(x\right)-2{cos}^3\left(x\right)}{cos\left(x\right)+sin\left(x\right)}dx$$$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{3sin\left(x\right)+cos\left(x\right)-2{sin}^3\left(x\right)-2{cos}^3\left(x\right)}{sin\left(x\right)+cos\left(x\right)}dx$$$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{3(sin\left(x\right)+cos\left(x\right))-2(sin\left(x\right)+cos\left(x\right))({sin}^2\left(x\right)+{cos}^2\left(x\right)-cos\left(x\right)sin\left(x\right))}{sin\left(x\right)+cos\left(x\right)}dx$$$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}3-2(1+sin\left(x\right)cos\left(x\right))dx$$$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}1-2sin\left(x\right)cos\left(x\right)dx={[x+{cos}^2\left(x\right)]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{\pi }{6}+\frac{1}{4}-\frac{3}{4}=\frac{\pi -3}{6}$$$$$$

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