Question

$$ \int_{0}^{\pi/4} (\tan^8 x + \tan^6 x) \, dx = ? $$

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{5}$
• D. $\frac{1}{7}$

Hint:

Reduction formula hint: $\int (\tan^n x + \tan^{n-2} x) \, dx = \frac{\tan^{n-1} x}{n-1}$. For $n=8$, the answer is simply $1/(8-1) = 1/7$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For definite integrals involving powers of $\tan x$, we look to factor out common powers to utilize the identity $1 + \tan^2 x = \sec^2 x$.

Step 2: Key Formula or Approach:
Factor out $\tan^6 x$ and apply the substitution $u = \tan x$.

Step 3: Detailed Explanation:
1. Factor the integrand: \[ \int_{0}^{\pi/4} \tan^6 x ( \tan^2 x + 1 ) \, dx \] 2. Apply the identity $1 + \tan^2 x = \sec^2 x$: \[ \int_{0}^{\pi/4} \tan^6 x \cdot \sec^2 x \, dx \] 3. Let $u = \tan x$, then $du = \sec^2 x \, dx$. 4. Change the limits of integration: When $x = 0$, $u = \tan(0) = 0$. When $x = \pi/4$, $u = \tan(\pi/4) = 1$. 5. Evaluate the integral: \[ \int_{0}^{1} u^6 \, du = \left[ \frac{u^7}{7} \right]_0^1 = \frac{1}{7} - 0 = \frac{1}{7} \]

Step 4: Final Answer
The value of the definite integral is $\frac{1}{7}$.