Question

in an attempt to calculate an integral, you perform a trigonometric substitution and are left with (int \frac{3sin^3(\theta)}{8cos^2(\theta)} d\theta). which of the following integrals did you start with? a. (int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx) b. (int \frac{3x^3}{8sqrt{1 - x^2}} dx) c. (int \frac{x^3}{sqrt{9 - x^2}} dx) d. (int \frac{sqrt{4 - x^2}}{2x^3} dx) e. (int \frac{x^3}{sqrt{9 - 4x^2}} dx)

Explanation:

Step1: Recall trigonometric substitution

For integrals involving $\sqrt{a^2 - b^2x^2}$, we use the substitution $x=\frac{a}{b}\sin\theta$ (or $x=\frac{a}{b}\cos\theta$). Let's analyze each option by reversing the substitution.

Step2: Analyze the given integral after substitution

We have $\int \frac{3\sin^3(\theta)}{8\cos^2(\theta)} d\theta$. Let's assume the original substitution was $x = \frac{3}{2}\sin\theta$ (so that $9 - 4x^2 = 9 - 4\times\frac{9}{4}\sin^2\theta = 9\cos^2\theta$, so $\sqrt{9 - 4x^2}=3\cos\theta$) or $x = r\sin\theta$ for some $r$.

Step3: Check Option E

Option E: $\int \frac{x^3}{\sqrt{9 - 4x^2}} dx$. Let $x=\frac{3}{2}\sin\theta$, then $dx=\frac{3}{2}\cos\theta d\theta$, and $\sqrt{9 - 4x^2}=\sqrt{9 - 4\times\frac{9}{4}\sin^2\theta}=3\cos\theta$. Substitute $x$ and $dx$ into the integral:

$x^3 = (\frac{3}{2}\sin\theta)^3=\frac{27}{8}\sin^3\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, $\sqrt{9 - 4x^2}=3\cos\theta$.

So the integral becomes:

$\int \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify numerator and denominator:

$\frac{27}{8}\times\frac{3}{2}\int \frac{\sin^3\theta}{\cos\theta}\times\cos\theta d\theta$ Wait, no, wait: Wait, $\sqrt{9 - 4x^2}=3\cos\theta$, so denominator is $3\cos\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, $x^3=\frac{27}{8}\sin^3\theta$.

So:

$\int \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify: $\frac{27}{8}\times\frac{3}{2}\times\frac{1}{3}\int \sin^3\theta d\theta$? Wait, no, let's recalculate:

Wait, $\frac{x^3}{\sqrt{9 - 4x^2}} dx = \frac{(\frac{3}{2}\sin\theta)^3}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

$= \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify: $\frac{27}{8}\times\frac{3}{2}\times\frac{1}{3}\sin^3\theta d\theta$? Wait, no, the $\cos\theta$ in the numerator (from $dx$) and denominator (from $\sqrt{9 - 4x^2}$) cancels? Wait, no:

Wait, $\sqrt{9 - 4x^2}=3\cos\theta$, so denominator is $3\cos\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, so:

$\frac{x^3}{\sqrt{9 - 4x^2}} dx = \frac{(\frac{3}{2}\sin\theta)^3}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify numerator: $(\frac{3}{2})^3\sin^3\theta = \frac{27}{8}\sin^3\theta$

Denominator: $3\cos\theta$

Multiply by $dx$: $\frac{27}{8}\sin^3\theta \times \frac{3}{2}\cos\theta d\theta \div (3\cos\theta)$

The $3\cos\theta$ in the denominator and the $\frac{3}{2}\cos\theta$ in the numerator: $\frac{3}{2}\cos\theta \div 3\cos\theta = \frac{1}{2}$

So overall: $\frac{27}{8} \times \frac{1}{2} \sin^3\theta d\theta = \frac{27}{16}\sin^3\theta d\theta$? Wait, that's not matching. Wait, maybe I made a mistake. Wait the given integral after substitution is $\frac{3\sin^3\theta}{8\cos^2\theta} d\theta$. Wait, maybe the substitution is $x = \frac{3}{2}\sin\theta$, so $9 - 4x^2 = 9 - 9\sin^2\theta = 9\cos^2\theta$, so $\sqrt{9 - 4x^2}=3\cos\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, $x^3 = (\frac{3}{2}\sin\theta)^3 = \frac{27}{8}\sin^3\theta$.

Wait, let's check Option E again: $\int \frac{x^3}{\sqrt{9 - 4x^2}} dx$

Substitute $x=\frac{3}{2}\sin\theta$:

$x^3 = \frac{27}{8}\sin^3\theta$, $\sqrt{9 - 4x^2}=3\cos\theta$, $dx=\frac{3}{2}\cos\theta d\theta$

So the integral becomes:

$\int \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify: $\frac{27}{8} \times \frac{3}{2} \times \frac{1}{3} \int \sin^3\theta d\theta$? Wait, no, the $3$ in the denominator and the $3$ in the numerator (from $\frac{3}{2}$) cancels? Wait, $\frac{27}{8}\sin^3\theta \times \frac{3}{2}\cos\theta d\theta$ divided by $3\cos\theta$:

$\frac{27}{8} \times \frac{3}{2} \times \frac{1}{\ 3} \sin^3\theta d\theta = \frac{27}{16}\sin^3\theta d\theta$? No, that's not matching. Wait maybe the substitution is $x = \frac{3}{2}\sin\theta$, but let's check Option A: $\int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx$

Let $x=\frac{3}{2}\sin\theta$, then $9 - 4x^2=9\cos^2\theta$, so $(9 - 4x^2)^{3/2}=27\cos^3\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, $x^3=\frac{27}{8}\sin^3\theta$

Substitute into Option A:

$\int \frac{2\times\frac{27}{8}\sin^3\theta}{27\cos^3\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify: $\frac{54}{8} \times \frac{3}{2} \times \frac{1}{27} \int \frac{\sin^3\theta}{\cos^3\theta} \times \cos\theta d\theta$

$\frac{54\times3}{8\times2\times27} \int \frac{\sin^3\theta}{\cos^2\theta} d\theta = \frac{162}{432} \int \frac{\sin^3\theta}{\cos^2\theta} d\theta = \frac{3}{8} \int \frac{\sin^3\theta}{\cos^2\theta} d\theta = \int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$

Yes! That matches the given integral after substitution. Wait, wait, I think I messed up the Option earlier. Wait Option A: $\int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx$

Let's redo the substitution for Option A:

Let $x = \frac{3}{2}\sin\theta$, so $dx = \frac{3}{2}\cos\theta d\theta$, $9 - 4x^2 = 9 - 4\times\frac{9}{4}\sin^2\theta = 9 - 9\sin^2\theta = 9\cos^2\theta$, so $(9 - 4x^2)^{3/2} = (9\cos^2\theta)^{3/2} = 27\cos^3\theta$ (since $\cos\theta$ is positive in the domain where substitution is valid, so we can take the positive root).

$x^3 = (\frac{3}{2}\sin\theta)^3 = \frac{27}{8}\sin^3\theta$

So the integrand becomes:

$\frac{2\times\frac{27}{8}\sin^3\theta}{27\cos^3\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify numerator: $2\times\frac{27}{8}\sin^3\theta \times \frac{3}{2}\cos\theta = \frac{54}{8}\times\frac{3}{2}\sin^3\theta\cos\theta = \frac{162}{16}\sin^3\theta\cos\theta$

Denominator: $27\cos^3\theta$

So the fraction is $\frac{162}{16\times27} \times \frac{\sin^3\theta\cos\theta}{\cos^3\theta} = \frac{162}{432} \times \frac{\sin^3\theta}{\cos^2\theta} = \frac{3}{8} \times \frac{\sin^3\theta}{\cos^2\theta}$

So the integral becomes $\int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$, which matches the given integral after substitution. Wait, but earlier I thought Option E, but no, Option A is correct? Wait no, wait the problem says "which of the following integrals did you start with" and after substitution we have $\int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$. Wait, maybe I made a mistake in Option A. Wait let's check again.

Wait Option A: $\int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx$

Substitution $x = \frac{3}{2}\sin\theta$:

$x = \frac{3}{2}\sin\theta \implies dx = \frac{3}{2}\cos\theta d\theta$

$9 - 4x^2 = 9 - 4(\frac{9}{4}\sin^2\theta) = 9 - 9\sin^2\theta = 9\cos^2\theta \implies (9 - 4x^2)^{3/2} = (9\cos^2\theta)^{3/2} = 27\cos^3\theta$ (since $\cos\theta \geq 0$ for the substitution to be valid, assuming $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$)

$x^3 = (\frac{3}{2}\sin\theta)^3 = \frac{27}{8}\sin^3\theta$

So the integrand:

$\frac{2 \times \frac{27}{8}\sin^3\theta}{27\cos^3\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify step by step:

First, numerator of the fraction: $2 \times \frac{27}{8}\sin^3\theta = \frac{54}{8}\sin^3\theta = \frac{27}{4}\sin^3\theta$

Denominator of the fraction: $27\cos^3\theta$

Multiply by $dx$: $\frac{27}{4}\sin^3\theta \times \frac{3}{2}\cos\theta d\theta \div 27\cos^3\theta$

Simplify:

$\frac{27 \times 3}{4 \times 2 \times 27} \times \frac{\sin^3\theta \cos\theta}{\cos^3\theta} d\theta = \frac{3}{8} \times \frac{\sin^3\theta}{\cos^2\theta} d\theta = \int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$

Yes! That's exactly the integral we have after substitution. So Option A is correct? Wait no, wait the original problem's after substitution integral is $\int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$, and when we substitute $x = \frac{3}{2}\sin\theta$ into Option A, we get that integral. So Option A is the correct answer? Wait but let's check Option E again. Wait maybe I made a mistake in Option E.

Wait Option E: $\int \frac{x^3}{\sqrt{9 - 4x^2}} dx$

Substitute $x = \frac{3}{2}\sin\theta$:

$x^3 = \frac{27}{8}\sin^3\theta$, $\sqrt{9 - 4x^2} = 3\cos\theta$, $dx = \frac{3}{2}\cos\theta d\theta$

So the integral becomes:

$\int \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta = \int \frac{27}{8} \times \frac{3}{2} \times \frac{1}{3} \sin^3\theta d\theta = \int \frac{27}{16}\sin^3\theta d\theta$, which is not the same as the given integral. So Option A is correct. Wait, but let's check the coefficients. The given after substitution has 3 in the numerator, 8 in the denominator. In Option A, the coefficient is 2 for $x^3$, and when we substitute, we get 3/8, which matches. So Option A is correct.

Wait, but let's check the options again. The options are:

A. $\int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx$

B. $\int \frac{3x^3}{8\sqrt{1 - x^2}} dx$

C. $\int \frac{x^3}{\sqrt{9 - x^2}} dx$

D. $\int \frac{\sqrt{4 - x^2}}{2x^3} dx$

E. $\int \frac{x^3}{\sqrt{9 - 4x^2}} dx$

So after substitution, the integral is $\int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$. Let's confirm the substitution for Option A:

Let $x = \frac{3}{2}\sin\theta \implies \sin\theta = \frac{2x}{3} \implies \theta = \arcsin(\frac{2x}{3})$

$dx = \frac{3}{2}\cos\theta d\theta$

$9 - 4x^2 = 9 - 4(\frac{9}{4}\sin^2\theta) = 9\cos^2\theta \implies (9 - 4x^2)^{3/2} = 27\cos^3\theta$

$x^3 = (\frac{3}{2}\sin\theta)^3 = \frac{27}{8}\sin^3\theta$

So the integrand:

$\frac{2 \times \frac{27}{8}\sin^3\theta}{27\cos^3\theta} \times \frac{3}{2}\cos\theta d\theta = \frac{54}{8} \times \frac{3}{2} \times \frac{1}{27} \times \frac{\sin^3\theta \cos\theta}{\cos^3\theta} d\theta$

Simplify:

$\frac{54 \times 3}{8 \times 2 \times 27} \times \frac{\sin^3\theta}{\cos^2\theta} d\theta = \frac{162}{432} \times \frac{\sin^3\theta}{\cos^2\theta} d\theta = \frac{3}{8} \times \frac{\sin^3\theta}{\cos^2\theta} d\theta = \int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$

Yes

Answer:

Step1: Recall trigonometric substitution

For integrals involving $\sqrt{a^2 - b^2x^2}$, we use the substitution $x=\frac{a}{b}\sin\theta$ (or $x=\frac{a}{b}\cos\theta$). Let's analyze each option by reversing the substitution.

Step2: Analyze the given integral after substitution

We have $\int \frac{3\sin^3(\theta)}{8\cos^2(\theta)} d\theta$. Let's assume the original substitution was $x = \frac{3}{2}\sin\theta$ (so that $9 - 4x^2 = 9 - 4\times\frac{9}{4}\sin^2\theta = 9\cos^2\theta$, so $\sqrt{9 - 4x^2}=3\cos\theta$) or $x = r\sin\theta$ for some $r$.

Step3: Check Option E

Option E: $\int \frac{x^3}{\sqrt{9 - 4x^2}} dx$. Let $x=\frac{3}{2}\sin\theta$, then $dx=\frac{3}{2}\cos\theta d\theta$, and $\sqrt{9 - 4x^2}=\sqrt{9 - 4\times\frac{9}{4}\sin^2\theta}=3\cos\theta$. Substitute $x$ and $dx$ into the integral:

$x^3 = (\frac{3}{2}\sin\theta)^3=\frac{27}{8}\sin^3\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, $\sqrt{9 - 4x^2}=3\cos\theta$.

So the integral becomes:

$\int \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify numerator and denominator:

$\frac{27}{8}\times\frac{3}{2}\int \frac{\sin^3\theta}{\cos\theta}\times\cos\theta d\theta$ Wait, no, wait: Wait, $\sqrt{9 - 4x^2}=3\cos\theta$, so denominator is $3\cos\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, $x^3=\frac{27}{8}\sin^3\theta$.

So:

$\int \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify: $\frac{27}{8}\times\frac{3}{2}\times\frac{1}{3}\int \sin^3\theta d\theta$? Wait, no, let's recalculate:

Wait, $\frac{x^3}{\sqrt{9 - 4x^2}} dx = \frac{(\frac{3}{2}\sin\theta)^3}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

$= \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify: $\frac{27}{8}\times\frac{3}{2}\times\frac{1}{3}\sin^3\theta d\theta$? Wait, no, the $\cos\theta$ in the numerator (from $dx$) and denominator (from $\sqrt{9 - 4x^2}$) cancels? Wait, no:

Wait, $\sqrt{9 - 4x^2}=3\cos\theta$, so denominator is $3\cos\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, so:

$\frac{x^3}{\sqrt{9 - 4x^2}} dx = \frac{(\frac{3}{2}\sin\theta)^3}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify numerator: $(\frac{3}{2})^3\sin^3\theta = \frac{27}{8}\sin^3\theta$

Denominator: $3\cos\theta$

Multiply by $dx$: $\frac{27}{8}\sin^3\theta \times \frac{3}{2}\cos\theta d\theta \div (3\cos\theta)$

The $3\cos\theta$ in the denominator and the $\frac{3}{2}\cos\theta$ in the numerator: $\frac{3}{2}\cos\theta \div 3\cos\theta = \frac{1}{2}$

So overall: $\frac{27}{8} \times \frac{1}{2} \sin^3\theta d\theta = \frac{27}{16}\sin^3\theta d\theta$? Wait, that's not matching. Wait, maybe I made a mistake. Wait the given integral after substitution is $\frac{3\sin^3\theta}{8\cos^2\theta} d\theta$. Wait, maybe the substitution is $x = \frac{3}{2}\sin\theta$, so $9 - 4x^2 = 9 - 9\sin^2\theta = 9\cos^2\theta$, so $\sqrt{9 - 4x^2}=3\cos\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, $x^3 = (\frac{3}{2}\sin\theta)^3 = \frac{27}{8}\sin^3\theta$.

Wait, let's check Option E again: $\int \frac{x^3}{\sqrt{9 - 4x^2}} dx$

Substitute $x=\frac{3}{2}\sin\theta$:

$x^3 = \frac{27}{8}\sin^3\theta$, $\sqrt{9 - 4x^2}=3\cos\theta$, $dx=\frac{3}{2}\cos\theta d\theta$

So the integral becomes:

$\int \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify: $\frac{27}{8} \times \frac{3}{2} \times \frac{1}{3} \int \sin^3\theta d\theta$? Wait, no, the $3$ in the denominator and the $3$ in the numerator (from $\frac{3}{2}$) cancels? Wait, $\frac{27}{8}\sin^3\theta \times \frac{3}{2}\cos\theta d\theta$ divided by $3\cos\theta$:

$\frac{27}{8} \times \frac{3}{2} \times \frac{1}{\ 3} \sin^3\theta d\theta = \frac{27}{16}\sin^3\theta d\theta$? No, that's not matching. Wait maybe the substitution is $x = \frac{3}{2}\sin\theta$, but let's check Option A: $\int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx$

Let $x=\frac{3}{2}\sin\theta$, then $9 - 4x^2=9\cos^2\theta$, so $(9 - 4x^2)^{3/2}=27\cos^3\theta$, $dx=\frac{3}{2}\cos\theta d\theta$, $x^3=\frac{27}{8}\sin^3\theta$

Substitute into Option A:

$\int \frac{2\times\frac{27}{8}\sin^3\theta}{27\cos^3\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify: $\frac{54}{8} \times \frac{3}{2} \times \frac{1}{27} \int \frac{\sin^3\theta}{\cos^3\theta} \times \cos\theta d\theta$

$\frac{54\times3}{8\times2\times27} \int \frac{\sin^3\theta}{\cos^2\theta} d\theta = \frac{162}{432} \int \frac{\sin^3\theta}{\cos^2\theta} d\theta = \frac{3}{8} \int \frac{\sin^3\theta}{\cos^2\theta} d\theta = \int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$

Yes! That matches the given integral after substitution. Wait, wait, I think I messed up the Option earlier. Wait Option A: $\int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx$

Let's redo the substitution for Option A:

Let $x = \frac{3}{2}\sin\theta$, so $dx = \frac{3}{2}\cos\theta d\theta$, $9 - 4x^2 = 9 - 4\times\frac{9}{4}\sin^2\theta = 9 - 9\sin^2\theta = 9\cos^2\theta$, so $(9 - 4x^2)^{3/2} = (9\cos^2\theta)^{3/2} = 27\cos^3\theta$ (since $\cos\theta$ is positive in the domain where substitution is valid, so we can take the positive root).

$x^3 = (\frac{3}{2}\sin\theta)^3 = \frac{27}{8}\sin^3\theta$

So the integrand becomes:

$\frac{2\times\frac{27}{8}\sin^3\theta}{27\cos^3\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify numerator: $2\times\frac{27}{8}\sin^3\theta \times \frac{3}{2}\cos\theta = \frac{54}{8}\times\frac{3}{2}\sin^3\theta\cos\theta = \frac{162}{16}\sin^3\theta\cos\theta$

Denominator: $27\cos^3\theta$

So the fraction is $\frac{162}{16\times27} \times \frac{\sin^3\theta\cos\theta}{\cos^3\theta} = \frac{162}{432} \times \frac{\sin^3\theta}{\cos^2\theta} = \frac{3}{8} \times \frac{\sin^3\theta}{\cos^2\theta}$

So the integral becomes $\int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$, which matches the given integral after substitution. Wait, but earlier I thought Option E, but no, Option A is correct? Wait no, wait the problem says "which of the following integrals did you start with" and after substitution we have $\int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$. Wait, maybe I made a mistake in Option A. Wait let's check again.

Wait Option A: $\int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx$

Substitution $x = \frac{3}{2}\sin\theta$:

$x = \frac{3}{2}\sin\theta \implies dx = \frac{3}{2}\cos\theta d\theta$

$9 - 4x^2 = 9 - 4(\frac{9}{4}\sin^2\theta) = 9 - 9\sin^2\theta = 9\cos^2\theta \implies (9 - 4x^2)^{3/2} = (9\cos^2\theta)^{3/2} = 27\cos^3\theta$ (since $\cos\theta \geq 0$ for the substitution to be valid, assuming $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$)

$x^3 = (\frac{3}{2}\sin\theta)^3 = \frac{27}{8}\sin^3\theta$

So the integrand:

$\frac{2 \times \frac{27}{8}\sin^3\theta}{27\cos^3\theta} \times \frac{3}{2}\cos\theta d\theta$

Simplify step by step:

First, numerator of the fraction: $2 \times \frac{27}{8}\sin^3\theta = \frac{54}{8}\sin^3\theta = \frac{27}{4}\sin^3\theta$

Denominator of the fraction: $27\cos^3\theta$

Multiply by $dx$: $\frac{27}{4}\sin^3\theta \times \frac{3}{2}\cos\theta d\theta \div 27\cos^3\theta$

Simplify:

$\frac{27 \times 3}{4 \times 2 \times 27} \times \frac{\sin^3\theta \cos\theta}{\cos^3\theta} d\theta = \frac{3}{8} \times \frac{\sin^3\theta}{\cos^2\theta} d\theta = \int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$

Yes! That's exactly the integral we have after substitution. So Option A is correct? Wait no, wait the original problem's after substitution integral is $\int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$, and when we substitute $x = \frac{3}{2}\sin\theta$ into Option A, we get that integral. So Option A is the correct answer? Wait but let's check Option E again. Wait maybe I made a mistake in Option E.

Wait Option E: $\int \frac{x^3}{\sqrt{9 - 4x^2}} dx$

Substitute $x = \frac{3}{2}\sin\theta$:

$x^3 = \frac{27}{8}\sin^3\theta$, $\sqrt{9 - 4x^2} = 3\cos\theta$, $dx = \frac{3}{2}\cos\theta d\theta$

So the integral becomes:

$\int \frac{\frac{27}{8}\sin^3\theta}{3\cos\theta} \times \frac{3}{2}\cos\theta d\theta = \int \frac{27}{8} \times \frac{3}{2} \times \frac{1}{3} \sin^3\theta d\theta = \int \frac{27}{16}\sin^3\theta d\theta$, which is not the same as the given integral. So Option A is correct. Wait, but let's check the coefficients. The given after substitution has 3 in the numerator, 8 in the denominator. In Option A, the coefficient is 2 for $x^3$, and when we substitute, we get 3/8, which matches. So Option A is correct.

Wait, but let's check the options again. The options are:

A. $\int \frac{2x^3}{(9 - 4x^2)^{3/2}} dx$

B. $\int \frac{3x^3}{8\sqrt{1 - x^2}} dx$

C. $\int \frac{x^3}{\sqrt{9 - x^2}} dx$

D. $\int \frac{\sqrt{4 - x^2}}{2x^3} dx$

E. $\int \frac{x^3}{\sqrt{9 - 4x^2}} dx$

So after substitution, the integral is $\int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$. Let's confirm the substitution for Option A:

Let $x = \frac{3}{2}\sin\theta \implies \sin\theta = \frac{2x}{3} \implies \theta = \arcsin(\frac{2x}{3})$

$dx = \frac{3}{2}\cos\theta d\theta$

$9 - 4x^2 = 9 - 4(\frac{9}{4}\sin^2\theta) = 9\cos^2\theta \implies (9 - 4x^2)^{3/2} = 27\cos^3\theta$

$x^3 = (\frac{3}{2}\sin\theta)^3 = \frac{27}{8}\sin^3\theta$

So the integrand:

$\frac{2 \times \frac{27}{8}\sin^3\theta}{27\cos^3\theta} \times \frac{3}{2}\cos\theta d\theta = \frac{54}{8} \times \frac{3}{2} \times \frac{1}{27} \times \frac{\sin^3\theta \cos\theta}{\cos^3\theta} d\theta$

Simplify:

$\frac{54 \times 3}{8 \times 2 \times 27} \times \frac{\sin^3\theta}{\cos^2\theta} d\theta = \frac{162}{432} \times \frac{\sin^3\theta}{\cos^2\theta} d\theta = \frac{3}{8} \times \frac{\sin^3\theta}{\cos^2\theta} d\theta = \int \frac{3\sin^3\theta}{8\cos^2\theta} d\theta$

Yes