How do you integrate $\int \frac{1}{\sqrt{x} \sqrt{4 - x}}$ $int 1/(sqrt[x]sqrt[4-x])$ ?

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How do you integrate $\int \frac{1}{\sqrt{x} \sqrt{4 - x}}$ $int 1/(sqrt[x]sqrt[4-x])$ ?

1 Answer

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Answer 1 · 2018-03-20T09:36:45

The answer is $= 2 arcsin (\frac{\sqrt{x}}{2}) + C$ $=2arcsin(sqrtx/2)+C$

Explanation:

Perform the substitution

$sin u = \frac{\sqrt{x}}{2}$ $sinu=sqrtx/2$ , $\Rightarrow$ $=>$ , $cos u d u = \frac{1}{4 \sqrt{x}} d x$ $cosudu=1/(4sqrtx)dx$

Therefore,

The integral is

$I = \int \frac{d x}{\sqrt{x} \sqrt{4 - x}} = \int \frac{4 \sqrt{x} cos u d u}{\sqrt{x} \sqrt{4 - 4 {sin}^{2} u}}$ $I=int(dx)/(sqrtxsqrt(4-x))=int(4sqrtxcosudu)/(sqrtxsqrt(4-4sin^2u))$

$= 2 \int 1 \cdot d u$ $=2int1*du$

$= 2 u$ $=2u$

$= 2 arcsin (\frac{\sqrt{x}}{2}) + C$ $=2arcsin(sqrtx/2)+C$

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How do you integrate $\int \frac{1}{\sqrt{x} \sqrt{4 - x}}$ $int 1/(sqrt[x]sqrt[4-x])$ ?

1 Answer

Explanation:

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How do you integrate ∫1√x√4−xint 1/(sqrt[x]sqrt[4-x])?

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How do you integrate $\int \frac{1}{\sqrt{x} \sqrt{4 - x}}$ $int 1/(sqrt[x]sqrt[4-x])$ ?