evaluate the following integral.\n int_{0}^{pi/3} \frac{7sin^{2}x}{sqrt{1 + cos x}}dx \n int_{0}^{pi/3}…

Explanation:

Step1: Use the double - angle formula

Recall that $\cos x = 2\cos^{2}\frac{x}{2}-1$, so $1 + \cos x=2\cos^{2}\frac{x}{2}$. Also, $\sin^{2}x = 1-\cos^{2}x=(1 - \cos x)(1 + \cos x)$. The integral $\int_{0}^{\frac{\pi}{3}}\frac{7\sin^{2}x}{\sqrt{1 + \cos x}}dx=7\int_{0}^{\frac{\pi}{3}}\frac{(1 - \cos x)(1 + \cos x)}{\sqrt{1 + \cos x}}dx=7\int_{0}^{\frac{\pi}{3}}(1 - \cos x)\sqrt{1 + \cos x}dx$. Since $\sqrt{1+\cos x}=\sqrt{2}\left|\cos\frac{x}{2}\right|$, and for $x\in[0,\frac{\pi}{3}]$, $\cos\frac{x}{2}\geq0$, so $\sqrt{1 + \cos x}=\sqrt{2}\cos\frac{x}{2}$. The integral becomes $7\int_{0}^{\frac{\pi}{3}}(1 - \cos x)\sqrt{2}\cos\frac{x}{2}dx$. Another way is to use the substitution $u = 1+\cos x$, then $du=-\sin xdx$. When $x = 0$, $u = 2$; when $x=\frac{\pi}{3}$, $u=\frac{3}{2}$. And $\sin^{2}x=1 - \cos x=1-(u - 1)=2 - u$. The integral $\int_{0}^{\frac{\pi}{3}}\frac{7\sin^{2}x}{\sqrt{1 + \cos x}}dx=- 7\int_{2}^{\frac{3}{2}}\frac{2 - u}{\sqrt{u}}du=7\int_{\frac{3}{2}}^{2}\frac{2 - u}{\sqrt{u}}du$.

Step2: Expand the integrand

Expand $\frac{2 - u}{\sqrt{u}}=\frac{2}{\sqrt{u}}-\sqrt{u}=2u^{-\frac{1}{2}}-u^{\frac{1}{2}}$. Then $7\int_{\frac{3}{2}}^{2}(2u^{-\frac{1}{2}}-u^{\frac{1}{2}})du=7\left[2\times\frac{u^{\frac{1}{2}}}{\frac{1}{2}}-\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{\frac{3}{2}}^{2}$.

Step3: Evaluate the definite - integral

$7\left[4u^{\frac{1}{2}}-\frac{2}{3}u^{\frac{3}{2}}\right]_{\frac{3}{2}}^{2}=7\left[\left(4\sqrt{2}-\frac{2}{3}\times2^{\frac{3}{2}}\right)-\left(4\sqrt{\frac{3}{2}}-\frac{2}{3}\times\left(\frac{3}{2}\right)^{\frac{3}{2}}\right)\right]$. $=7\left[\left(4\sqrt{2}-\frac{4\sqrt{2}}{3}\right)-\left(2\sqrt{6}-\frac{2}{3}\times\frac{3\sqrt{6}}{2\sqrt{2}}\right)\right]$. $=7\left[\frac{8\sqrt{2}}{3}-\left(2\sqrt{6}-\frac{\sqrt{6}}{\sqrt{2}}\right)\right]$. $=7\left[\frac{8\sqrt{2}}{3}-\left(2\sqrt{6}-\sqrt{3}\right)\right]$. $=7\left[\frac{8\sqrt{2}}{3}-2\sqrt{6}+\sqrt{3}\right]=\frac{56\sqrt{2}}{3}-14\sqrt{6} + 7\sqrt{3}$.

Answer:

$\frac{56\sqrt{2}}{3}-14\sqrt{6}+7\sqrt{3}$