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JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Indefinite Integrals

Let

f (x) = \int \frac{dx}{( 3 + 4 x ^{2} ) 4 - 3 x ^{2}}, ∣ x ∣ < \frac{2}{3}

. If

f (0) = 0

and

f (1) = \frac{1}{α β} tan^{- 1} (\frac{α}{β}), α, β > 0

, then

α^{2} + β^{2}

is equal to ____

Answer: 28

Solution

Let

x = \frac{1}{t}

, so

d x = - \frac{1}{t ^{2}} d t

.

f (x) = \int \frac{- d t / t ^{2}}{\frac{3 t ^{2} + 4}{t ^{2}} \cdot \frac{4 t ^{2} - 3}{t}} = - \int \frac{d t}{( 3 t ^{2} + 4 ) 4 t ^{2} - 3}

Let

4 t^{2} - 3 = z^{2}

, so

8 t d t = 2 z d z

, giving

t d t = \frac{z d z}{4}

.

Also,

t^{2} = \frac{z ^{2} + 3}{4}

, so:

f (x) = - \frac{1}{4} \int \frac{z d z}{( 3 \cdot \frac{z ^{2} + 3}{4} + 4 ) z} = - \frac{1}{4} \int \frac{d z}{\frac{3 z ^{2} + 25}{4}} = - \frac{1}{3} \int \frac{d z}{z ^{2} + \frac{25}{3}}

= - \frac{1}{5 3} tan^{- 1} (\frac{3 z}{5}) + C = - \frac{1}{5 3} tan^{- 1} (\frac{3}{5} \frac{4 - 3 x ^{2}}{x ^{2}}) + C

Using

f (0) = 0

: as

x \to 0

, the argument

\to \infty

, so

tan^{- 1} (\cdot) \to \frac{π}{2}

, giving

C = \frac{π}{10 3}

.

f (1) = - \frac{1}{5 3} tan^{- 1} (\frac{3}{5}) + \frac{π}{10 3} = \frac{1}{5 3} [\frac{π}{2} - tan^{- 1} (\frac{3}{5})] = \frac{1}{5 3} tan^{- 1} (\frac{5}{3})

Comparing with

\frac{1}{α β} tan^{- 1} (\frac{α}{β})

:

α = 5

,

β = 3

.

Therefore:

α^{2} + β^{2} = 25 + 3 = 28

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Integer answer

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