Understanding the Tangent Integral

The problem asks us to calculate the sum \(I_8 + I_6\) where \(I_n\) is defined as the definite integral:

$$I_n = \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^n \theta \ d\theta$$

To find the sum \(I_8 + I_6\), we can first look for a relationship between consecutive integrals in the sequence, specifically \(I_n\) and \(I_{n-2}\).

Deriving the Reduction Formula

Let's consider the sum \(I_n + I_{n-2}\):

• Write out the sum using the definition of \(I_n\): $$I_n + I_{n-2} = \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^n \theta \ d\theta + \int_0^{\tfrac{\pi}{4}} \tan^{n-2} \theta \ d\theta$$
• Combine the integrals since they have the same limits: $$I_n + I_{n-2} = \displaystyle\int_0^{\tfrac{\pi}{4}} (\tan^n \theta + \tan^{n-2} \theta) \ d\theta$$
• Factor out the common term \(\tan^{n-2} \theta\): $$I_n + I_{n-2} = \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^{n-2} \theta (\tan^2 \theta + 1) \ d\theta$$
• Use the trigonometric identity: \(\tan^2 \theta + 1 = \sec^2 \theta\). $$I_n + I_{n-2} = \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^{n-2} \theta \sec^2 \theta \ d\theta$$
• Now, we use substitution. Let \(u = \tan \theta\). Then, the differential is \(du = \sec^2 \theta \ d\theta\). We also need to change the limits of integration:
  • When \(\theta = 0\), \(u = \tan(0) = 0\).
  • When \(\theta = \tfrac{\pi}{4}\), \(u = \tan(\tfrac{\pi}{4}) = 1\).
• Substitute \(u\) and \(du\) into the integral: $$I_n + I_{n-2} = \displaystyle\int_0^1 u^{n-2} \ du$$
• Evaluate the integral with respect to \(u\): $$I_n + I_{n-2} = \left[ \dfrac{u^{(n-2)+1}}{(n-2)+1} \right]_0^1 = \left[ \dfrac{u^{n-1}}{n-1} \right]_0^1$$ (This is valid for \(n-1 \neq 0\), i.e., \(n \neq 1\))
• Apply the limits: $$I_n + I_{n-2} = \dfrac{1^{n-1}}{n-1} - \dfrac{0^{n-1}}{n-1} = \dfrac{1}{n-1}$$

So, we have found the reduction formula:

$$I_n + I_{n-2} = \dfrac{1}{n-1}$$

Calculating I8 + I6

We need to find the value of \(I_8 + I_6\). We can use the reduction formula derived above by setting \(n=8\).

Substitute \(n=8\) into the formula \(I_n + I_{n-2} = \dfrac{1}{n-1}\):

$$I_8 + I_{6} = \dfrac{1}{8-1}$$

$$I_8 + I_{6} = \dfrac{1}{7}$$

Therefore, the value of \(I_8 + I_6\) is \(\frac{1}{7}\).