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Sagot :
Answer:
The result of the integral is:
[tex]\arcsin{(\frac{x}{3})} + C[/tex]
Step-by-step explanation:
We are given the following integral:
[tex]\int \frac{dx}{\sqrt{9-x^2}}[/tex]
Trigonometric substitution:
We have the term in the following format: [tex]a^2 - x^2[/tex], in which a = 3.
In this case, the substitution is given by:
[tex]x = a\sin{\theta}[/tex]
So
[tex]dx = a\cos{\theta}d\theta[/tex]
In this question:
[tex]a = 3[/tex]
[tex]x = 3\sin{\theta}[/tex]
[tex]dx = 3\cos{\theta}d\theta[/tex]
So
[tex]\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}[/tex]
We have the following trigonometric identity:
[tex]\sin^{2}{\theta} + \cos^{2}{\theta} = 1[/tex]
So
[tex]1 - \sin^{2}{\theta} = \cos^{2}{\theta}[/tex]
Replacing into the integral:
[tex]\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C[/tex]
Coming back to x:
We have that:
[tex]x = 3\sin{\theta}[/tex]
So
[tex]\sin{\theta} = \frac{x}{3}[/tex]
Applying the arcsine(inverse sine) function to both sides, we get that:
[tex]\theta = \arcsin{(\frac{x}{3})}[/tex]
The result of the integral is:
[tex]\arcsin{(\frac{x}{3})} + C[/tex]
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