`int_(0)^(pi//4)(tan^(4)x + tan^(2)x)dx=`

Options

A. 1
B. `(1)/(2)`
C. `(1)/(3)`
D. `(1)/(4)`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{4}} \left( \tan^4 x + \tan^2 x \right) dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We can factor out \( \tan^2 x \) from the integrand:
\[ \int_{0}^{\frac{\pi}{4}} \left( \tan^4 x + \tan^2 x \right) dx = \int_{0}^{\frac{\pi}{4}} \tan^2 x \left( \tan^2 x + 1 \right) dx \]
Since \( \tan^2 x + 1 = \sec^2 x \), we can rewrite the integral as:
\[ \int_{0}^{\frac{\pi}{4}} \tan^2 x \sec^2 x \, dx \]

### Step 2: Substitution
Let \( t = \tan x \). Then, the derivative \( dt = \sec^2 x \, dx \). Changing the limits:
- When \( x = 0 \), \( t = \tan(0) = 0 \)
- When \( x = \frac{\pi}{4} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \)

Thus, the integral becomes:
\[ \int_{0}^{1} t^2 \, dt \]

### Step 3: Evaluate the Integral
Now we can evaluate the integral:
\[ \int_{0}^{1} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

### Final Answer
Therefore, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{4}} \left( \tan^4 x + \tan^2 x \right) dx = \frac{1}{3} \]

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