Yes because 1/3 and 2/6 are the same. For example: 2/4 and 1/2 are the same because 2/4 can be reduced down to 1/2. In your equation 2/6 can be reduced to 1/2. Therefore 2/6 = 1/3
Yes, if you were to divide 1/3 you would get 1/6. Now since you divided 1/3 into two parts to get 1/6, now if you would like to get 1/3 out then you must double 1/6 into 2/6 which would be equal to 1/3
I hope this made sense, I know you wanted a more simple yes or no but i believe if you explain to someone how they got the answer than you wouldn't need help next time. xx
![I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\ = [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\ =0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\ ](https://tex.z-dn.net/?f=I%3D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5En%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%28x%29%2Asin%5E%7Bn-1%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D%20%5B-cos%28x%29%2Asin%5E%7Bn-1%7D%28x%29%5D_0%5E%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2B%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bcos%28x%29%2Asin%5E%7Bn-2%7D%28x%29%2Acos%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D0%20%2B%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bcos%5E2%28x%29%2Asin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7B%281-sin%5E2%28x%29%29%2Asin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20-%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5En%28x%29%20%5C%2C%20dx%5C%5C%0A%0A)
![\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\ = \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\ = \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\ ](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5E%7B3%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%5Cfrac%7B2%7D%7B3%7D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%5Cdfrac%7B2%7D%7B3%7D%5C%20%5B-cos%28x%29%5D_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%3D%5Cdfrac%7B2%7D%7B3%7D%20%5C%5C%0A%0A%0A%0A%0A)
