29\int \sqrt{cos^{2n -1}x}(\sqrt{1 - cos^{2}x}dx = \int \sqrt{cos^{2n -1}x}(sinx)dx..take cosx = t..since it's a symmetric function so it can be written as \int_{-\pi /2}^{\pi /2}{} \sqrt{cos^{2n -1}x}(sinx)dx = 2\int_{0}^{\pi /2}{} \sqrt{cos^{2n -1}x}(sinx)dx
take cosx = t
\Rightarrow -\int_{1}^{0}{t^{n - 1/2}}dt = -\left(\frac{t^{n+1/2}}{n+1/2} \right)
Now put the limits ..u will get the answer..
