For $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, if $y(x)=\int \frac{\csc x+\sin x}{\csc x \sec x+\tan x \sin^{2} x} dx$ and $\lim_{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0$, then $y\left(\frac{\pi}{4}\right)$ is equal to

(D) $\frac{1}{\sqrt{2}} \tan^{-1}\left(-\frac{1}{2}\right)$

(A) $\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$

(B) $\frac{1}{2} \tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$

(C) $-\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Indefinite Integrals — JEE Main · 2024 · 29 Jan 2024 · Shift 1 Question with Solution

For

x \in (- \frac{π}{2}, \frac{π}{2})

, if

y (x) = \int \frac{c s c x + s i n x}{c s c x s e c x + t a n x s i n ^{2} x} d x

and

lim_{x \to (\frac{π}{2})^{-}} y (x) = 0

, then

y (\frac{π}{4})

is equal to

Answer: (D)

y (x) = \int \frac{csc x + sin x}{csc x sec x + tan x sin ^{2} x} d x

Simplify the denominator:

csc x sec x + tan x sin^{2} x = \frac{1}{sin x cos x} + \frac{sin ^{3} x}{cos x} = \frac{1 + sin ^{4} x}{sin x cos x}

.

The numerator is

csc x + sin x = \frac{1 + sin ^{2} x}{sin x}

y (x) = \int \frac{( 1 + sin ^{2} x ) cos x}{1 + sin ^{4} x} d x

Let

t = sin x

y = \int \frac{1 + t ^{2}}{1 + t ^{4}} d t = \int \frac{1 + \frac{1}{t ^{2}}}{t ^{2} + \frac{1}{t ^{2}}} d t

With

u = t - \frac{1}{t}

d u = (1 + \frac{1}{t ^{2}}) d t

and

t^{2} + \frac{1}{t ^{2}} = u^{2} + 2

y = \int \frac{d u}{u ^{2} + 2} = \frac{1}{2} tan^{- 1} (\frac{u}{2}) + C = \frac{1}{2} tan^{- 1} (\frac{t - 1/ t}{2}) + C

x \to (π /2)^{-}

t \to 1^{-}

, so

t - 1/ t \to 0

and

y \to C

. Given the limit is 0,

C = 0

.

At

x = π /4

t = \frac{1}{2}

, so

t - \frac{1}{t} = \frac{1}{2} - 2 = - \frac{1}{2}

, and

\frac{u}{2} = - \frac{1}{2}

y (\frac{π}{4}) = \frac{1}{2} tan^{- 1} (- \frac{1}{2})

Key confirmed: D.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option