If c o s e c 5 x d x = cot x c o s e c x ( c o s e c 2 x + 3 2 ) + log e | tan x 2 | + C , where and C is the constant of integration, then the value of 8 equals _______ . [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If $\int c o s e c^{5} x d x = α \cot x c o s e c x (c o s e c^{2} x + \frac{3}{2}) + β \log_{e} | \tan \frac{x}{2} | + C,$ where $α, β \in R$ and $C$ is the constant of integration, then the value of $8 (α + β)$ equals _______ . [2024]

Ans.
(1)

$I_{n} = \int c o s e c^{n} x d x$

Using reduction formula, we get

$I_{n} = \frac{- c o s e c^{n - 2} x \cot x}{n - 1} + \frac{n - 2}{n - 1} I_{n - 2}$

$∴$ $I_{5} = \int c o s e c^{5} x d x$

$= \frac{- c o s e c^{3} x \cot x}{4} + \frac{3}{4} I_{3} + C$

$= \frac{- c o s e c^{3} x \cot x}{4} + \frac{3}{4} [\frac{- c o s e c x \cot x}{2} + \frac{1}{2} \int c o s e c x d x] + C$

$= \frac{- 1}{4} c o s e c^{3} x \cot x - \frac{3}{8} c o s e c x \cot x + \frac{3}{8} \log_{e} | \tan \frac{x}{2} | + C$

$= \frac{- 1}{4} c o s e c x \cot x [c o s e c^{2} x + \frac{3}{2}] + \frac{3}{8} \log_{e} | \tan \frac{x}{2} | + C$

$= α \cot x c o s e c x [c o s e c^{2} x + \frac{3}{2}] + β \log_{e} | \tan \frac{x}{2} | + C (∵ Given)$

On comparing, we get $α = \frac{- 1}{4}$ and $β = \frac{3}{8}$

Hence, $8 (α + β) = 8 (\frac{- 1}{4} + \frac{3}{8}) = 8 \times \frac{1}{8} = 1$

Want Us to Email you About Special Offers & Updates?