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

Definite Integrals — JEE Main · 2025 · 22 Jan 2025 · Shift 1 Question with Solution

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) dx

and

I_{2} = \int_{0}^{π /4} x f (x) dx

. Then

7 I_{1} + 12 I_{2}

is equal to :

Answer: (C)

Factor

f (x)

f (x) = 7 tan^{8} x + 7 tan^{6} x - 3 tan^{4} x - 3 tan^{2} x = (7 tan^{6} x - 3 tan^{2} x) (1 + tan^{2} x) = (7 tan^{6} x - 3 tan^{2} x) sec^{2} x .

For

I_{1}

, substitute

t = tan x

d t = sec^{2} x d x

, with limits

0 \to 1

I_{1} = \int_{0}^{π /4} (7 tan^{6} x - 3 tan^{2} x) sec^{2} x d x = \int_{0}^{1} (7 t^{6} - 3 t^{2}) d t = [t^{7} - t^{3}]_{0}^{1} = 0.

For

I_{2}

, integrate by parts with

u = x

and

d v = (7 tan^{6} x - 3 tan^{2} x) sec^{2} x d x

, so

v = tan^{7} x - tan^{3} x

I_{2} = [x (tan^{7} x - tan^{3} x)]_{0}^{π /4} - \int_{0}^{π /4} (tan^{7} x - tan^{3} x) d x .

x = π /4

tan x = 1

, so the bracket is

x (1 - 1) = 0

; at

x = 0

it is also

0

. So the boundary term vanishes.

I_{2} = - \int_{0}^{π /4} tan^{3} x (tan^{4} x - 1) d x = - \int_{0}^{π /4} tan^{3} x (tan^{2} x - 1) (tan^{2} x + 1) d x = - \int_{0}^{π /4} (tan^{5} x - tan^{3} x) sec^{2} x d x .

Substituting

t = tan x

again:

I_{2} = - \int_{0}^{1} (t^{5} - t^{3}) d t = - [\frac{t ^{6}}{6} - \frac{t ^{4}}{4}]_{0}^{1} = - (\frac{1}{6} - \frac{1}{4}) = \frac{1}{12} .

7 I_{1} + 12 I_{2} = 7 (0) + 12 (\frac{1}{12}) = 1.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option