If \(\rm I_n = \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^n \theta \ d\theta \), then I8 + I6 equals:
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 \(\frac{1}{7}\).
Important Questions from Definite Integrals
The Legendre polynomials P n(x), n = 0, 1, 2, ..., satisfying the orthogonailty condition \(\int_{{\rm{ - 1}}}^{\rm{1}} {{{\rm{P}}_{\rm{n}}}\left( {\rm{x}} \right){{\rm{P}}_{\rm{m}}}} \left( {\rm{x}} \right){\rm{dx}}\,{\rm{ = }}\,\frac{{\rm{2}}}{{{\rm{2n + 1}}}}{{\rm{\delta }}_{{\rm{nm}}}}\) on the interval [-1, +1], may be defined by the Rodrigues formula P n(x) = \(\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}{\rm{n!}}}}\frac{{{{\rm{d}}^{\rm{n}}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 1}}} \right)^{\rm{n}}}\) . The value of the definite integral \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \) is
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