Carroll Lewis
05/16/2024 · Primary School
\( \int \frac { d x } { 4 - x ^ { 2 } \sqrt { 4 - x ^ { 2 } } } \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the integral
\[
I = \int \frac{dx}{4 - x^2 \sqrt{4 - x^2}},
\]
we can start by making a substitution to simplify the expression. Let's use the substitution
\[
x = 2 \sin \theta.
\]
Then, we have
\[
dx = 2 \cos \theta \, d\theta,
\]
and we can rewrite \(4 - x^2\) as follows:
\[
4 - x^2 = 4 - (2 \sin \theta)^2 = 4 - 4 \sin^2 \theta = 4(1 - \sin^2 \theta) = 4 \cos^2 \theta.
\]
Now, we also need to compute \(\sqrt{4 - x^2}\):
\[
\sqrt{4 - x^2} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta.
\]
Substituting these into the integral, we get:
\[
I = \int \frac{2 \cos \theta \, d\theta}{4 - (2 \sin \theta)^2 (2 \cos \theta)} = \int \frac{2 \cos \theta \, d\theta}{4 - 4 \sin^2 \theta (2 \cos \theta)}.
\]
This simplifies to:
\[
I = \int \frac{2 \cos \theta \, d\theta}{4(1 - \sin^2 \theta \cos \theta)} = \int \frac{2 \cos \theta \, d\theta}{4 \cos^2 \theta} = \int \frac{1}{2} \, d\theta = \frac{1}{2} \theta + C.
\]
Now, we need to convert back to \(x\). Recall that
\[
\sin \theta = \frac{x}{2} \implies \theta = \arcsin\left(\frac{x}{2}\right).
\]
Thus, we have:
\[
I = \frac{1}{2} \arcsin\left(\frac{x}{2}\right) + C.
\]
Therefore, the final result for the integral is:
\[
\int \frac{dx}{4 - x^2 \sqrt{4 - x^2}} = \frac{1}{2} \arcsin\left(\frac{x}{2}\right) + C.
\]
Quick Answer
The integral \( \int \frac { d x } { 4 - x ^ { 2 } \sqrt { 4 - x ^ { 2 } } } \) simplifies to \( \frac{1}{2} \arcsin\left(\frac{x}{2}\right) + C \).
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