How do you use substitution to integrate $\frac{d x}{{(1 - {(8 x + 4)}^{2})}^{\frac{1}{2}}}$ $dx / (1− ( 8 x+4 ) ^2 )^ (1/2)$ ?

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How do you use substitution to integrate $\frac{d x}{{(1 - {(8 x + 4)}^{2})}^{\frac{1}{2}}}$ $dx / (1− ( 8 x+4 ) ^2 )^ (1/2)$ ?

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Answer 1 · 2018-04-05T05:28:15

The answer is $= \frac{1}{8} arcsin (8 x + 4) + C$ $=1/8arcsin(8x+4)+C$

Explanation:

Perform the substitution

$sin θ = 8 x + 4$ $sintheta=8x+4$ , $\Rightarrow$ $=>$ , $cos θ d θ = 8 d x$ $costhetad theta=8dx$

$\sqrt{1 - {(8 s + 4)}^{2}} = \sqrt{1 - {sin}^{2} θ} = cos θ$ $sqrt(1-(8s+4)^2)=sqrt(1-sin^2theta)=costheta$

Therefore, the integral is

$\int \frac{d x}{\sqrt{1 - {(8 x + 4)}^{2}}} = \frac{1}{8} \int \frac{cos θ d θ}{cos θ}$ $int(dx)/sqrt(1-(8x+4)^2)=1/8int(costheta d theta)/(cos theta)$

$= \frac{1}{8} θ$ $=1/8theta$

$= \frac{1}{8} arcsin (8 x + 4) + C$ $=1/8arcsin(8x+4)+C$

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How do you use substitution to integrate $\frac{d x}{{(1 - {(8 x + 4)}^{2})}^{\frac{1}{2}}}$ $dx / (1− ( 8 x+4 ) ^2 )^ (1/2)$ ?

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Explanation:

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