NIMCET 2013 — Mathematics PYQ

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Question

NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

If $I_n = \int_{0}^{\frac{\pi}{4}} \tan^n \theta \, d\theta$ , then $I_8 + I_6$ equals:

Choose the correct answer:

Explanation

Solution

Step 1: Express the sum in integral form

Given $I_n = \int_{0}^{\frac{\pi}{4}} \tan^n \theta \, d\theta$ . We need to find $I_8 + I_6$ .

I_8 + I_6 = \int_{0}^{\frac{\pi}{4}} \tan^8 \theta \, d\theta + \int_{0}^{\frac{\pi}{4}} \tan^6 \theta \, d\theta

Step 2: Combine the integrals

Since the limits of integration are the same, we can combine them:

I_8 + I_6 = \int_{0}^{\frac{\pi}{4}} (\tan^8 \theta + \tan^6 \theta) \, d\theta

Step 3: Simplify the integrand

Factor out $\tan^6 \theta$ :

I_8 + I_6 = \int_{0}^{\frac{\pi}{4}} \tan^6 \theta (1 + \tan^2 \theta) \, d\theta

Using the trigonometric identity $1 + \tan^2 \theta = \sec^2 \theta$ :

I_8 + I_6 = \int_{0}^{\frac{\pi}{4}} \tan^6 \theta \sec^2 \theta \, d\theta

Step 4: Solve the integral using substitution

Let $u = \tan \theta$ , then $du = \sec^2 \theta \, d\theta$ .

Changing the limits:

When $\theta = 0$ , $u = \tan(0) = 0$ .
When $\theta = \frac{\pi}{4}$ , $u = \tan(\frac{\pi}{4}) = 1$ .

Now, the integral becomes:

I_8 + I_6 = \int_{0}^{1} u^6 \, du

I_8 + I_6 = \left[ \frac{u^{6+1}}{6+1} \right]_{0}^{1}

I_8 + I_6 = \left[ \frac{u^7}{7} \right]_{0}^{1}

I_8 + I_6 = \frac{1^7}{7} - \frac{0^7}{7} = \frac{1}{7}

General Formula Note:

In general, for $I_n = \int_{0}^{\frac{\pi}{4}} \tan^n \theta \, d\theta$ , the reduction relation is $I_n + I_{n-2} = \frac{1}{n-1}$ .

Correct Option: 4. $\frac{1}{7}$

Choose the correct answer:

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