Question
Evaluate the following integrals:
$\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
$\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
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Answer
Let $\text{I}=\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{I}=\int\sqrt{4-\text{t}^2}\text{dt}$
$=\int\sqrt{2^2-\text{t}^2}\text{dt}$
$=\frac{\text{t}^2}{2}\sqrt{2^2-\text{t}^2}+\frac{4}{2}\sin^{-1}\frac{\text{t}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\sin\text{x}\sqrt{4-\sin^2\text{x}}+2\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{I}=\int\sqrt{4-\text{t}^2}\text{dt}$
$=\int\sqrt{2^2-\text{t}^2}\text{dt}$
$=\frac{\text{t}^2}{2}\sqrt{2^2-\text{t}^2}+\frac{4}{2}\sin^{-1}\frac{\text{t}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\sin\text{x}\sqrt{4-\sin^2\text{x}}+2\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$
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