Evaluate an integral using trigonometric substitution

Q: What substitution is best for this integral?

Use 2x = 3 sin(theta), because it turns sqrt(9 - 4x^2) into 3 cos(theta) and simplifies the radical.

Q: What is the final antiderivative?

The antiderivative is -sqrt(9 - 4x^2)/x - 2arcsin(2x/3) + C.

Question

1.) $\int \frac{\sqrt{(9-4x^2)}\,dx}{x^2}$

Solution:

$u = 2x; \quad a = 3$

Let $u = a\sin	heta$

$\Rightarrow 2x = 3\sin	heta$

$\Rightarrow x = \frac{3}{2}\sin	heta$

$\Rightarrow dx = \frac{3}{2}\cos	heta\, d	heta$

And $\sin	heta = 2x/3$. This means that the side opposite to angle $	heta$ has measure $2x$, hypotenuse has measure 3, and the adjacent side to angle $	heta$ has measure $\sqrt{(9-4x^2)}$.

Therefore,

$\int \frac{\sqrt{(9-4x^2)}\,dx}{x^2}$

$= \int \frac{\sqrt{(9-9\sin^2	heta)}\,(3/2\cos	heta\, d	heta)}{9/4\sin^2	heta}$

$= \int \frac{\sqrt{(9)(1-\sin^2	heta)}\,(3/2\cos	heta\, d	heta)}{9/4\sin^2	heta}$

$= \int \frac{3\cos	heta\,(3/2\cos	heta\, d	heta)}{9/4\sin^2	heta}$

$= \int \frac{2\cos^2	heta\, d	heta}{\sin^2	heta}$

$= \int 2\cot^2	heta\, d	heta$

$= 2\int (\csc^2	heta - 1)\, d	heta$

$= -2\cot	heta - 2	heta$

Now, substitute values of $\cot	heta$ and $	heta$ with the values of the reference triangle as mentioned above. Thus,

$= -2\sqrt{(9-4x^2)}/2x - 2\arcsin(2x/3) + C$

$= -1/x\,\sqrt{(9-4x^2)} - 2\arcsin(2x/3) + C \quad 	ext{Ans.}$

Acemy AI · Accepted Answer

> **Key concept**: This integral is a good match for a trig substitution because of the square root $\sqrt{9-4x^2}$. The only tricky part is keeping the algebra clean after the substitution.

### Step by step
We want to compute

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

### Step 1: Choose a substitution
Because of the form $9-4x^2$, let

$$2x=3\sin	heta,$$

so that

$$x=\frac32\sin	heta,\qquad dx=\frac32\cos	heta\,d	heta.$$

Also,

$$\sqrt{9-4x^2}=\sqrt{9-9\sin^2	heta}=3\cos	heta.$$

### Step 2: Rewrite the integral
Substitute everything in:

$$\int \frac{\sqrt{9-4x^2}}{x^2}\,dx
=\int \frac{3\cos	heta}{\left(\frac32\sin	hetaight)^2}\cdot \frac32\cos	heta\,d	heta.$$

Since

$$\left(\frac32\sin	hetaight)^2=\frac94\sin^2	heta,$$

we get

$$\int \frac{3\cos	heta}{\frac94\sin^2	heta}\cdot \frac32\cos	heta\,d	heta
=\int 2\frac{\cos^2	heta}{\sin^2	heta}\,d	heta
=\int 2\cot^2	heta\,d	heta.$$

### Step 3: Integrate
Use

$$\cot^2	heta=\csc^2	heta-1,$$

so

$$\int 2\cot^2	heta\,d	heta
=2\int (\csc^2	heta-1)\,d	heta
=-2\cot	heta-2	heta+C.$$

### Step 4: Convert back to $x$
From the triangle relation,

$$\sin	heta=\frac{2x}{3}$$

so

$$	heta=\arcsin\left(\frac{2x}{3}ight).$$

Also,

$$\cot	heta=\frac{\cos	heta}{\sin	heta}
=\frac{\sqrt{9-4x^2}/3}{2x/3}
=\frac{\sqrt{9-4x^2}}{2x}.$$

Therefore,

$$-2\cot	heta-2	heta+C
=-2\cdot \frac{\sqrt{9-4x^2}}{2x}-2\arcsin\left(\frac{2x}{3}ight)+C.$$

So the antiderivative is

$$\boxed{-\frac{\sqrt{9-4x^2}}{x}-2\arcsin\left(\frac{2x}{3}ight)+C}.$$

### Pitfall alert
The most common slip here is losing a factor of 2 or 3 when converting $x=\frac32\sin	heta$ and $dx=\frac32\cos	heta\,d	heta$. Another easy mistake is forgetting that $\sqrt{9-9\sin^2	heta}=3\cos	heta$ assumes the usual right-triangle setup with $	heta$ chosen so that $\cos	heta\ge 0$.

### Try different conditions
If the integrand were $\int \frac{\sqrt{9-4x^2}}{x}\,dx$, the same substitution would still work, but the algebra would simplify differently and the final answer would involve a mix of logarithmic and inverse-trig terms. If the square root were $\sqrt{a^2-b^2x^2}$ instead, the substitution pattern would be the same: set $bx=a\sin	heta$.

### Further reading
trigonometric substitution, inverse trigonometric function, antiderivative

Question

Evaluate an integral using trigonometric substitution

Expert Verified Solution

Step by step

Step 1: Choose a substitution

Step 2: Rewrite the integral

FAQ

What substitution is best for this integral?

What is the final antiderivative?