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Sagot :
It looks like the integral might be
[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx[/tex]
or perhaps
[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx[/tex]
Take note of the fact that both integrands are defined only over the interval -3 < x < 3.
For either integral, we substitute x = 3 sin(θ) and dx = 3 cos(θ) dθ.
Note that we want this substitution to be reversible, so we must restrict -π/2 ≤ θ ≤ π/2, an interval over which sine has an inverse. Then θ = arcsin(x/3).
The first case then reduces to
[tex]\displaystyle \int (9 - (3\sin(\theta))^2)^{3/2} (3 \cos(\theta) \, d\theta) = 3 \times 9^{3/2} \int (1 - \sin^2(\theta))^{3/2} \cos(\theta) \, d\theta \\\\ = 81 \int (\cos^2(\theta))^{3/2} \cos(\theta) \, d\theta \\\\ = 81 \int |\cos^3(\theta)| \cos(\theta) \, d\theta[/tex]
By definition of absolute value,
[tex]\displaystyle 81 \int |\cos^3(\theta)| \cos(\theta) \, d\theta = \begin{cases}\displaystyle 81 \int \cos^4(\theta) \, d\theta & \text{if }\cos(\theta) \ge 0 \\ \displaystyle -81 \int \cos^4(\theta) \, d\theta & \text{if }\cos(\theta) < 0\end{cases}[/tex]
and these cases correspond to 0 ≤ θ < π/2 and π/2 < θ ≤ π, respectively. But we are assuming -π/2 ≤ θ ≤ π/2, so the negative case doesn't matter to us.
You can compute the remaining antiderivative by exploiting the half-angle identity for cosine,
[tex]\cos^2(\theta) = \dfrac{1 + \cos(2\theta)}2[/tex]
Then
[tex]\cos^4(\theta) = \left(\cos^2(\theta)\right)^2 = \dfrac{1 + 2\cos(2\theta) + \cos^2(2\theta)}4 = \dfrac{3 + 4\cos(2\theta) + \cos(4\theta)}8[/tex]
and so
[tex]\displaystyle \int \cos^4(\theta) \, d\theta = \dfrac{12\theta + 8\sin(2\theta) + \sin(4\theta)}{32} + C[/tex]
We can simplify this using the double angle identity for (co)sine,
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = 1 - 2 sin²(θ)
as well as the relations,
sin(arcsin(x/3)) = x/3
cos(arcsin(x/3)) = √(9 - x²)/3
which gives us
[tex]\displaystyle \int \cos^4(\theta) \, d\theta = \dfrac{12\theta + 16 \sin(\theta) \cos(\theta) + 4 \sin(\theta) \cos(\theta) (1 - 2\sin^2(\theta))}{32} + C[/tex]
Putting this in terms of x, we get
[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx \\ = 81 \times \dfrac{12\arcsin\left(\frac x3\right) + 16 \times \frac x3 \times \frac{\sqrt{9-x^2}}3 + 4\times\frac x3\times\frac{\sqrt{9-x^2}}3 \left(1 - 2\left(\frac x3\right)^2\right)}{32} + C[/tex]
[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx = 81 \times \dfrac{12\arcsin\left(\frac x3\right) + \frac{16x\sqrt{9-x^2}}9 + \frac{4x(9-2x^2)\sqrt{9-x^2}}{81}}{32} + C[/tex]
[tex]\boxed{\displaystyle \int (9 - x^2)^{3/2} \, dx = \dfrac{12\arcsin\left(\frac x3\right) + (180x-8x^3)\sqrt{9-x^2}}{32} + C}[/tex]
If you were asking about the other integral, the first few steps are similar and you end up with the far more trivial integral and antiderivative
[tex]\displaystyle \frac19 \int \frac{d\theta}{\cos^2(\theta)} = \frac19 \int \sec^2(\theta) \, d\theta = \frac19 \tan(\theta) + C[/tex]
Putting it back in terms of x, we get
[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac19 \tan\left(\arcsin\left(\frac x3\right)\right) + C[/tex]
Recall that tan(θ) = sin(θ)/cos(θ), so
[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac19 \times \frac{\frac x3}{\frac{\sqrt{9-x^2}}3} + C[/tex]
[tex]\boxed{\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac{x}{9\sqrt{9-x^2}} + C}[/tex]
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