Question

If $I_n = \int_0^{\pi/4} \tan^n \theta d\theta$, then $I_{12} + I_{10}$ is equal to}

• A. $1/8$
• B. $1/12$
• C. $1/11$
• D. $1/10$

Hint:

For integrals of the form $\int_0^{\pi/4} \tan^n x dx$, a useful recurrence relation is $I_n + I_{n-2} = \frac{1}{n-1}$. This property can quickly solve such problems by direct substitution of $n$.

Answer 1

The Correct Option is A

Step 1: Establish the recurrence relation for $I_n$. We are given $I_n = \int_0^{\pi/4} \tan^n \theta d\theta$. Let's consider the sum $I_n + I_{n-2}$. \[ I_n + I_{n-2} = \int_0^{\pi/4} (\tan^n \theta + \tan^{n-2} \theta) d\theta \] \[ I_n + I_{n-2} = \int_0^{\pi/4} \tan^{n-2} \theta (\tan^2 \theta + 1) d\theta \] Using the trigonometric identity $\tan^2 \theta + 1 = \sec^2 \theta$: \[ I_n + I_{n-2} = \int_0^{\pi/4} \tan^{n-2} \theta \sec^2 \theta d\theta \]
Step 2: Evaluate the integral. To evaluate this integral, let $u = \tan \theta$. Then $du = \sec^2 \theta d\theta$. When $\theta = 0$, $u = \tan(0) = 0$. When $\theta = \pi/4$, $u = \tan(\pi/4) = 1$. So the integral becomes: \[ I_n + I_{n-2} = \int_0^1 u^{n-2} du \] \[ I_n + I_{n-2} = \left[ \frac{u^{n-1{n-1} \right]_0^1 \]
Step 3: Apply the limits of integration. Substitute the limits of integration: \[ I_n + I_{n-2} = \frac{1^{n-1{n-1} - \frac{0^{n-1{n-1} \] \[ I_n + I_{n-2} = \frac{1}{n-1} \]
Step 4: Use the relation for the given problem. We need to find $I_{12} + I_{10}$. This corresponds to setting $n = 12$ in the recurrence relation. \[ I_{12} + I_{12-2} = \frac{1}{12-1} \] \[ I_{12} + I_{10} = \frac{1}{11} \] Thus, the value of $I_{12} + I_{10}$ is $1/11$.