Let In = ∫ tanⁿ x dx, (n 1). If I₄ + I₆ = a tan⁵ x + bx⁵ + C, where C is a constant of integration

Q: Wait, why do we add $I_n$ and $I_{n-2}$? How would I know to do that in an exam?

That is a brilliant question! In JEE Advanced, you often encounter integrals of trigonometric functions with high powers. When you see powers differing by two, like $n$ and $n-2$, it is a massive hint to use the identity $1 + \tan^2 x = \sec^2 x$. By grouping them, you create a $\sec^2 x$ term, which is the derivative of $\tan x$. It is a classic technique to simplify complex integrals into something manageable.

Q: Why is the $x^5$ term coefficient $b=0$? Did I miss an integration constant?

You did not miss anything! The constant of integration $C$ is already accounted for in the problem statement. The term $bx^5$ is a distractor. Since our integration result is $\frac{1}{5} \tan^5 x + C$, there is no $x^5$ term present. Therefore, its coefficient $b$ must be zero. Always trust your derivation over the form of the expression given in the question!

Q: Can I solve this using integration by parts instead of substitution?

You certainly could, but it would be much longer and more prone to calculation errors. Integration by parts is powerful, but for this specific reduction formula, the substitution $u = \tan x$ is the most elegant and direct path. Always look for the 'derivative-function' pair before jumping into more complex methods.

Question

Let $I_n = \int 	an^n x \, dx, (n > 1)$. If $I_4 + I_6 = a 	an^5 x + bx^5 + C$, where C is a constant of integration, then the ordered pair $(a, b)$ is equal to:

Options:

(A) $(-\frac{1}{5}, 1)$

(B) $(\frac{1}{5}, 0)$

(C) $(\frac{1}{5}, -1)$

(D) $(-\frac{1}{5}, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the threshold of a complex integral. You see $I_n = \int 	an^n x \, dx$, and your first instinct might be to panic.

In the world of JEE Advanced, these problems are not designed to break you; they are designed to reveal a hidden, beautiful structure. Today, we are going to peel back the layers of this integral and discover why the sum $I_4 + I_6$ is a masterpiece of algebraic simplification.

### Pattern Recognition

We are given $I_n = \int 	an^n x \, dx$. The problem asks us to evaluate $I_4 + I_6$.

Notice the gap between the powers: $6 - 4 = 2$. This is not a coincidence. Whenever you see powers of trigonometric functions differing by two, your mind should immediately jump to the Pythagorean identity.

Let us write out the sum:
$$I_n + I_{n-2} = \int 	an^n x \, dx + \int 	an^{n-2} x \, dx$$

By the linearity of the integral, we can combine these into one:
$$\int (	an^n x + 	an^{n-2} x) \, dx$$

### The Algebraic Dance

Now, look at the integrand: $	an^n x + 	an^{n-2} x$. There is a common factor here, the lower power, $	an^{n-2} x$.

Let us factor it out:
$$\int 	an^{n-2} x (	an^2 x + 1) \, dx$$

Suddenly, the expression inside the parenthesis, $	an^2 x + 1$, screams at us. It is the fundamental identity $\sec^2 x$. Our integral transforms into:
$$\int 	an^{n-2} x \cdot \sec^2 x \, dx$$

This is the moment of clarity. We have a function, $	an x$, and its derivative, $\sec^2 x$, sitting right there.

### The Substitution Magic

This is where the magic happens. Let us set $u = 	an x$. Then, the differential $du$ is simply $\sec^2 x \, dx$.

Our integral becomes:
$$\int u^{n-2} \, du$$

This is a simple power rule application. The integral is:
$$\frac{u^{n-1}}{n-1} + C$$

Substituting back $u = 	an x$, we get the general reduction formula:
$$I_n + I_{n-2} = \frac{	an^{n-1} x}{n-1} + C$$

### The Final Comparison

Now, let us apply this to our specific case, $I_4 + I_6$. By setting $n = 6$, we get:
$$I_6 + I_4 = \frac{	an^{6-1} x}{6-1} + C = \frac{1}{5} 	an^5 x + C$$

The problem states that $I_4 + I_6 = a 	an^5 x + bx^5 + C$. By comparing our result with this expression, we see that $a = \frac{1}{5}$ and $b = 0$.

The $x^5$ term was merely a test of your confidence in the derivation. You have successfully navigated the trap and arrived at the correct ordered pair **$(a, b) = (\frac{1}{5}, 0)$**.

Animated Solution for Mathematics - Indefinite Integration: Let $I_{n} = \int tan^{n} x d x, (n > 1)$ . If $I_{4} + I_{6} = a tan^{5} x + b x^{5} + C$ , where C is a constant of integration, then the ordered pair $(a, b)$ is equal to:

Select Answer:

Visualized Solution

Understanding the Goal

Setting up $I_{n} + I_{n - 2}$

Factoring the Integrand

Applying Trigonometric Identity

Integration by Substitution

Executing the Substitution

Integrating in terms of $u$

Reverting to $x$ (General Formula)

Applying to $I_{4} + I_{6}$

Comparing Coefficients

Final Conclusion

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

Pattern Recognition

The Algebraic Dance

The Substitution Magic

The Final Comparison

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