Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
$\int\limits^{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
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Answer
Let $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(1-\cos^2\text{x}\big)}(-\tan^2\text{x})\cos^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(\sin^2\text{x})}\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}(\sin\text{x})\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}\big(1-\cos^2\text{x}\big)\sin\text{x dx}$ $\Big(\sin\text{x}=\sin\text{x}\text{ for }0\leq\text{x}\leq\frac{\pi}{2}\Big)$
Put $\cos\text{x}=\text{z}^2$
$\therefore\ -\sin\text{x dx}=2\text{z dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow0$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow0$
$\therefore\ \text{I}=-\int^\limits0_1\text{z}(1-\text{z}^4)2\text{z dz}$
$=-2\int^0\limits_1\text{z}^2\text{ dz}+2\int^\limits0_1\text{z}^6\text{ dz}$
$=-2\times\Big[\frac{\text{z}^3}{3}\Big]^0_1+2\times\Big[\frac{\text{z}^7}{7}\Big]^0_1$
$=-\frac{2}{3}\big(0-1\big)+\frac{2}{7}\big(0-1\big)$
$=\frac{2}{3}-\frac{2}{74}$
$=\frac{8}{21}$
$=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(1-\cos^2\text{x}\big)}(-\tan^2\text{x})\cos^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(\sin^2\text{x})}\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}(\sin\text{x})\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}\big(1-\cos^2\text{x}\big)\sin\text{x dx}$ $\Big(\sin\text{x}=\sin\text{x}\text{ for }0\leq\text{x}\leq\frac{\pi}{2}\Big)$
Put $\cos\text{x}=\text{z}^2$
$\therefore\ -\sin\text{x dx}=2\text{z dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow0$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow0$
$\therefore\ \text{I}=-\int^\limits0_1\text{z}(1-\text{z}^4)2\text{z dz}$
$=-2\int^0\limits_1\text{z}^2\text{ dz}+2\int^\limits0_1\text{z}^6\text{ dz}$
$=-2\times\Big[\frac{\text{z}^3}{3}\Big]^0_1+2\times\Big[\frac{\text{z}^7}{7}\Big]^0_1$
$=-\frac{2}{3}\big(0-1\big)+\frac{2}{7}\big(0-1\big)$
$=\frac{2}{3}-\frac{2}{74}$
$=\frac{8}{21}$
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