COMPASSLearning

Talk to us Register Login

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Definite Integrals

The value of the integral

\int_{0}^{\frac{1}{2}} \frac{1 + 3}{( ( x + 1 ) ^{2} ( 1 - x ) ^{6} ) ^{\frac{1}{4}}} d x

is ____.

Answer: 2

Solution

Let

x = sin θ

, then:

I = \int_{0}^{π /6} \frac{( 1 + 3 ) sec ^{2} ( θ /2 )}{( tan ( θ /2 ) - 1 ) ^{2}} d θ

Substituting

y = tan (θ /2)

:

I = \int_{0}^{2 - 3} \frac{2 ( 1 + 3 )}{( y - 1 ) ^{2}} d y = 2 (1 + 3) [- \frac{1}{y - 1}]_{0}^{2 - 3}

= 2 (1 + 3) (\frac{3 - 1}{2}) = 2

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · 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 →