COMPASSLearning

Talk to us Register Login

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Definite Integrals

If

I = \frac{2}{π} \int_{- π /4}^{π /4} \frac{d x}{( 1 + e ^{s i n x} ) ( 2 - cos 2 x )}

then

27 I^{2}

equals ____

Answer: 4

Solution

Using King's rule (property of definite integrals):

2 I = \frac{2}{π} \int_{- π /4}^{π /4} [\frac{1}{( 1 + e ^{s i n x} ) ( 2 - cos 2 x )} + \frac{1}{( 1 + e ^{- s i n x} ) ( 2 - cos 2 x )}] dx

Simplifying:

I = \frac{1}{π} \int_{- π /4}^{π /4} \frac{d x}{2 - cos 2 x} = \frac{2}{π} \int_{0}^{π /4} \frac{d x}{2 - cos 2 x}

Substituting

cos 2 x = 1 - 2 sin^{2} x

:

I = \frac{2}{π} \int_{0}^{π /4} \frac{sec ^{2} x d x}{1 + 3 tan ^{2} x} = \frac{2}{3 π} [tan^{- 1} (3 tan x)]_{0}^{π /4} = \frac{2}{3 3}

Therefore:

27 I^{2} = 27 \times \frac{4}{27} = 4

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer

Practise more: all JEE Advanced PYQs on Definite Integrals →

More JEE Advanced questions on Definite Integrals

The value of the definite integral
$\int_{0}^{2} \frac{1}{3 ^{x} + 3} d x is$
JEE Advanced 2026View solution →
If
$α = \int_{\frac{1}{2}}^{2} \frac{tan ^{- 1} x}{2 x ^{2} - 3 x + 2} d x$

then the value of $7 tan (\frac{2 α 7}{π})$ is ____.
(Here, the inverse trigonometric function $tan^{- 1} x$ assumes values in …
JEE Advanced 2025View solution →
Let the function $f : [1, \infty) \to R$ be defined by …
JEE Advanced 2024View solution →
Let $f : [0, \frac{π}{2}] \to [0, 1]$ be the function defined by $f (x) = sin^{2} x$ and let $g : [0, \frac{π}{2}] \to [0, \infty)$ be the function defined by $g (x) = \frac{π x}{2} - x^{2}$ .

The value of …
JEE Advanced 2024View solution →
Let $f : [0, \frac{π}{2}] \to [0, 1]$ be the function defined by $f (x) = sin^{2} x$ and let $g : [0, \frac{π}{2}] \to [0, \infty)$ be the function defined by $g (x) = \frac{π x}{2} - x^{2}$ .

The value of …
JEE Advanced 2024View solution →
For $x \in R$ , let $tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2})$ . Then the minimum value of the function $f : R \to R$ defined by …
JEE Advanced 2023View solution →