Question
Evaluate the following definite integrals:
$\int_{0}^\limits{4}\frac{1}{\sqrt{4\text{x}-\text{x}^2}}\text{ dx}$
$\int_{0}^\limits{4}\frac{1}{\sqrt{4\text{x}-\text{x}^2}}\text{ dx}$
✓
Answer
We have,
$\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4\text{x}-\text{x}^2}}$
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4-4+4\text{x}-\text{x}^2}}$ [Add and subtract 4 in denominator]
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4-(\text{x}^2-4\text{x}+4)}}$
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{(2)^2-(\text{x}-2)^2}}$
$=\Big[\sin^{-1}\Big(\frac{\text{x}-2}{2}\Big)\Big]^4_0$ $\bigg[\because\int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]$
$=\sin^{-1}(1)-\sin^{-1}(-1)$
$=\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)$
$=\frac{2\pi}{2}=\pi$
$\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4\text{x}-\text{x}^2}}$
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4-4+4\text{x}-\text{x}^2}}$ [Add and subtract 4 in denominator]
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4-(\text{x}^2-4\text{x}+4)}}$
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{(2)^2-(\text{x}-2)^2}}$
$=\Big[\sin^{-1}\Big(\frac{\text{x}-2}{2}\Big)\Big]^4_0$ $\bigg[\because\int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]$
$=\sin^{-1}(1)-\sin^{-1}(-1)$
$=\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)$
$=\frac{2\pi}{2}=\pi$
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