What is `int_(-pi//4)^(pi//4) tan^3 x dx` equal to

Options

A. `sqrt3`
B. `1/3`
C. `1/2`
D. 0

Answer

To solve the integral \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^3 x \, dx \), we can use the property of definite integrals that states:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

### Step 1: Apply the property of definite integrals

We have \( a = -\frac{\pi}{4} \) and \( b = \frac{\pi}{4} \). Thus, \( a + b = 0 \). Therefore, we can rewrite the integral as:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^3 x \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^3\left(0 - x\right) \, dx \]

### Step 2: Simplify the integral

Using the property of the tangent function, we know that:

\[ \tan(-x) = -\tan(x) \]

Thus, we can write:

\[ \tan^3(-x) = (-\tan(x))^3 = -\tan^3(x) \]

So, we can express the integral as:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} -\tan^3 x \, dx \]

### Step 3: Combine the two expressions for \( I \)

Now we have two expressions for \( I \):

1. \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^3 x \, dx \)
2. \( I = -\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^3 x \, dx \)

Let’s denote the second expression as \( -I \). Therefore, we have:

\[ I = -I \]

### Step 4: Solve for \( I \)

Adding \( I \) to both sides gives:

\[ I + I = 0 \implies 2I = 0 \implies I = 0 \]

### Conclusion

Thus, the value of the integral is:

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^3 x \, dx = 0 \]

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