Question Number 68434 by mhmd last updated on 10/Sep/19
Answered by mind is power last updated on 10/Sep/19
![∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx let f(x)=((3sin(x)−2sin^2 (x))/(sin(x)+cos(x))) ∫_(π/6) ^(π/3) f(x)dx=∫_(π/6) ^(π/3) f((π/2)−x)dx=∫_(π/6) ^(π/3) ((3cos(x)−2cos^3 (x))/(sin(x)+cos(x)))dx 2∫_(π/6) ^(π/3) f(x)dx=∫_(π/6) ^(π/3) ((3sin(x)−2sin^2 (x))/(cos(x)+sin(x)))dx+∫_(π/6) ^(π/3) ((3cos(x)−2cos^3 (x))/(cos(x)+sin(x)))dx =∫_(π/6) ^(π/3) ((3sin(x)+cos(x)−2sin^3 (x)−2cos^3 (x))/(sin(x)+cos(x)))dx =∫_(π/6) ^(π/3) ((3(sin(x)+cos(x))−2(sin(x)+cos(x))(sin^2 (x)+cos^2 (x)−cos(x)sin(x)))/(sin(x)+cos(x)))dx =∫_(π/6) ^(π/3) 3−2(1+sin(x)cos(x))dx =∫_(π/6) ^(π/3) 1−2sin(x)cos(x)dx=[x+cos^2 (x)]_(π/6) ^(π/3) =(π/6)+(1/4)−(3/4)=((π−3)/6)](Q68439.png)
$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+\mathrm{b}−\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{3sin}\left(\mathrm{x}\right)−\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{cos}\left(\mathrm{x}\right)} \\ $$$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} {f}\left({x}\right){dx}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} {f}\left(\frac{\pi}{\mathrm{2}}−{x}\right){dx}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{3}{cos}\left({x}\right)−\mathrm{2}{cos}^{\mathrm{3}} \left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx} \\ $$$$\mathrm{2}\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} {f}\left({x}\right){dx}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{3}{sin}\left({x}\right)−\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right)}{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx}+\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{3}{cos}\left({x}\right)−\mathrm{2}{cos}^{\mathrm{3}} \left({x}\right)}{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx} \\ $$$$=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{3}{sin}\left({x}\right)+{cos}\left({x}\right)−\mathrm{2}{sin}^{\mathrm{3}} \left({x}\right)−\mathrm{2}{cos}^{\mathrm{3}} \left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx} \\ $$$$=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{3}\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)−\mathrm{2}\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)\left({sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)−{cos}\left({x}\right){sin}\left({x}\right)\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx} \\ $$$$=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \mathrm{3}−\mathrm{2}\left(\mathrm{1}+{sin}\left({x}\right){cos}\left({x}\right)\right){dx} \\ $$$$=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \mathrm{1}−\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right){dx}=\left[{x}+{cos}^{\mathrm{2}} \left({x}\right)\right]_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} =\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}=\frac{\pi−\mathrm{3}}{\mathrm{6}} \\ $$$$ \\ $$