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MCQMediumJEE 2025Integration Techniques (Substitution, Parts, Partial Fractions)

JEE Mathematics 2025 Question with Solution

Let for f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x, I1=∫0π4f(x)dxI_1 = \int_0^{\frac{\pi}{4}} f(x)dx and I2=∫0π4xf(x)dxI_2 = \int_0^{\frac{\pi}{4}} x f(x)dx. Then 7I1+12I27I_1 + 12I_2 is equal to:

  • A

    2π2\pi

  • B

    π\pi

  • C

    11

  • D

    22

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x I1=∫0π4f(x) dx,I2=∫0π4xf(x) dxI_1 = \int_0^{\frac{\pi}{4}} f(x) \, dx, \qquad I_2 = \int_0^{\frac{\pi}{4}} x f(x) \, dx

Find: 7I1+12I27I_1 + 12I_2

For I1I_1, let t=tan⁡xt = \tan x. Then dt=sec⁡2x dxdt = \sec^2 x \, dx and the expression reduces as shown in the solution to

I1=∫01(7t6−3t2) dtI_1 = \int_0^1 (7t^6 - 3t^2) \, dt

Hence,

I1=[t7−t3]01=1−1=0I_1 = [t^7 - t^3]_0^1 = 1 - 1 = 0

Now for I2I_2,

I2=∫0π/4x(7tan⁡6x−3tan⁡2x)(sec⁡2x) dxI_2 = \int_0^{\pi/4} x (7 \tan^6 x - 3 \tan^2 x) (\sec^2 x) \, dx

Using integration by parts as shown,

I2=[x(tan⁡7x−tan⁡3x)]0π/4−∫0π/4(tan⁡7x−tan⁡3x) dxI_2 = [x (\tan^7 x - \tan^3 x)]_0^{\pi/4} - \int_0^{\pi/4} (\tan^7 x - \tan^3 x) \, dx =0−∫0π/4tan⁡3x(tan⁡4x−1) dx= 0 - \int_0^{\pi/4} \tan^3 x (\tan^4 x - 1) \, dx =0−∫0π/4tan⁡3x(tan⁡2x−1)(1+tan⁡2x) dx= 0 - \int_0^{\pi/4} \tan^3 x (\tan^2 x - 1) (1 + \tan^2 x) \, dx

Again put t=tan⁡xt = \tan x. Then

I2=−∫01(t5−t3) dtI_2 = - \int_0^1 (t^5 - t^3) \, dt =−[t66−t44]01= - \left[ \frac{t^6}{6} - \frac{t^4}{4} \right]_0^1 =−(16−14)=112= - \left( \frac{1}{6} - \frac{1}{4} \right) = \frac{1}{12}

Therefore,

7I1+12I2=7(0)+12(112)=17I_1 + 12I_2 = 7(0) + 12 \left( \frac{1}{12} \right) = 1

So, the correct option is C.

Symmetry Substitution

Given: the same integrals I1I_1 and I2I_2. Find: 7I1+12I27I_1 + 12I_2.

Use the substitution x↦π4−xx \mapsto \frac{\pi}{4} - x. The solution states that after tangent addition or reduction formulas,

f ⁣(π4−x)f\!\Big(\frac{\pi}{4}-x\Big)

can be combined with f(x)f(x) to produce a simpler expression.

Then consider

J:=∫0π/4(7+12x)f(x) dx=7I1+12I2J := \int_0^{\pi/4} (7 + 12x) f(x) \, dx = 7I_1 + 12I_2

Replacing xx by π4−x\frac{\pi}{4} - x and adding the two forms gives a telescoping logarithmic expression. As stated in the solution, all endpoint singularities cancel and the remaining finite terms simplify exactly to 11.

Therefore, 7I1+12I2=17I_1 + 12I_2 = 1, so the correct option is C.

Common mistakes

  • Using t=tan⁡xt = \tan x but forgetting that dt=sec⁡2x dxdt = \sec^2 x \, dx. This loses an essential factor and changes the integral incorrectly. Always convert both the trigonometric power and the differential together.

  • Trying to integrate I2I_2 directly without integration by parts. The factor xx suggests integration by parts after recognizing a derivative pattern involving tan⁡x\tan x. Use the antiderivative of the remaining trigonometric part as the second factor.

  • Expanding powers of tan⁡x\tan x unnecessarily in I1I_1. The expression simplifies neatly after the substitution shown in the solution. Look for factorization with 1+tan⁡2x=sec⁡2x1+\tan^2 x = \sec^2 x instead of brute-force expansion.

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