Let for f(x) = 7 tan⁸x + 7 tan⁶x - 3 tan⁴x - 3 tan²x, I_1 = ∫_0^{π/4} f(x)dx and I_2 = ∫_0^{π/4} xf(

Q: Wait, why does I_1 evaluate to zero? Did I make a calculation error?

That is a fantastic question! It is not an error; it is a beautiful property of the function over that specific interval. When you integrate (7t^6 - 3t^2) from 0 to 1, you get (1^7 - 1^3) - (0 - 0), which is 1 - 1 = 0. It is a moment of mathematical serendipity where the areas above and below the axis cancel out perfectly. Trust your algebra!

Q: Why do we use Integration by Parts (IBP) for I_2?

We use IBP because we have a product of an algebraic function (x) and a function whose integral we already know (f(x)). The ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) suggests setting u = x (Algebraic) and dv = f(x)dx (Trigonometric). It is the most efficient way to reduce the complexity of the integral.

Q: Is there a shortcut to avoid the long calculation for I_2?

In JEE Advanced, while there are sometimes shortcuts, the path of IBP is the most reliable and rigorous method here. The 'shortcut' is realizing that the boundary term vanishes, which saves you a lot of time. Always look for those boundary terms to simplify to zero!

Question

Let for $f(x)=7~	an^{8}x+7~	an^{6}x-3~	an^{4}x-3~	an^{2}x$, $I_{1}=\int_{0}^{\pi/4}f(x)dx$ and $I_{2}=\int_{0}^{\pi/4}xf(x)dx$. Then $7I_{1}+12I_{2}$ is equal to:

Options:

(A) 2

(B) 1

(C) $2\pi$

(D) $\pi$

Sigma Percentile · Accepted Answer

Welcome, future IITian! Today, we are going to dismantle a problem that looks like a nightmare but is actually a masterclass in algebraic elegance.

### Analyzing the Setup

When you first look at $f(x) = 7	an^8x + 7	an^6x - 3	an^4x - 3	an^2x$, it is natural to feel a surge of intimidation. It looks like a chaotic mess of powers.

But in the world of JEE Advanced, the most complex-looking expressions are often just hiding a simple secret. The secret here is **grouping**.

If we group the first two terms and the last two terms, we get:
$$f(x) = 7	an^6x(	an^2x + 1) - 3	an^2x(	an^2x + 1)$$

Suddenly, the fundamental identity $	an^2x + 1 = \sec^2x$ emerges from the shadows. The function simplifies beautifully to:
$$f(x) = (7	an^6x - 3	an^2x)\sec^2x$$

### Solving for $I_1$

This is the turning point of our journey. Now, for $I_1 = \int_0^{\pi/4} f(x) dx$, we use the substitution $t = 	an x$.

The integral transforms into:
$$\int_0^1 (7t^6 - 3t^2) dt$$

Integrating this is a simple application of the power rule:
$$\left[ t^7 - t^3 ight]_0^1$$

When we plug in the limits, we get $1 - 1 = 0$. Yes, the area is zero! It is a moment of pure mathematical harmony.

### Solving for $I_2$

Now, we move to the second act: $I_2 = \int_0^{\pi/4} x f(x) dx$. We have an algebraic function $x$ multiplied by our complex function $f(x)$.

This is the classic setup for **Integration by Parts**. We set $u = x$ and $dv = f(x) dx$. We already know the integral of $f(x)$ from our work on $I_1$, so we have our $v$ ready to go!

The boundary term $\left[ x(	an^7x - 	an^3x) ight]_0^{\pi/4}$ vanishes because at $x = \pi/4$, $	an x = 1$, making the term $1 - 1 = 0$. We are left with the integral of the difference, which simplifies to $\frac{1}{12}$.

### Final Calculation

Finally, we combine our results:
$$7(0) + 12\left(\frac{1}{12}ight) = 1$$

The complexity melts away, leaving us with a clean, elegant integer. The final answer is **1**.

This is the beauty of calculus—no matter how scary the problem looks, if you stay calm and follow the logic, the answer will reveal itself.

Animated Solution for Mathematics - Definite Integration: Let for $f (x) = 7 tan^{8} x + 7 tan^{6} x - 3 tan^{4} x - 3 tan^{2} x$ , $I_{1} = \int_{0}^{π /4} f (x) d x$ and $I_{2} = \int_{0}^{π /4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to:

Select Answer:

Visualized Solution

Analyzing $f (x)$

Factorizing $f (x)$

Evaluating $I_{1}$

The Value of $I_{1}$

Setting up $I_{2}$ with IBP

Boundary Term of $I_{2}$

Simplifying the $I_{2}$ Integral

Final Integration for $I_{2}$

The Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Solving for $I_{1}$

Solving for $I_{2}$

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Factorizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

The Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​ with IBP

Boundary Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​ Integral

Final Integration for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

The Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​