The value of e - 4 + 0 4 e - x tan 50 x ? d x 0 4 e - x ( tan 49 x + tan 51 x ) ? d x is: [2023]

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.
$The value of \frac{e^{- \frac{π}{4}} + \int_{0}^{\frac{π}{4}} e^{- x} \tan^{50} x d x}{\int_{0}^{\frac{π}{4}} e^{- x} (\tan^{49} x + \tan^{51} x) d x} is:$ [2023]

1 51

2 49

3 25

4 50

Ans.
(4)

$\frac{e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x}$

First, simplify, $\int_{0}^{π / 4} e^{- x} \tan^{50} x d x$

$= {[- e^{- x} {(\tan x)}^{50}]}_{0}^{π / 4} + \int_{0}^{π / 4} e^{- x} (50) \tan^{49} x \cdot \sec^{2} x d x$

$= - e^{- π / 4} + 0 + 50 \int_{0}^{π / 4} e^{- x} {(\tan x)}^{49} (1 + \tan^{2} x) d x$

$= - e^{- π / 4} + 50 \int_{0}^{π / 4} e^{- x} [{(\tan x)}^{51} + {(\tan x)}^{49}] d x$

Now, $\frac{- e^{- π / 4} + \int_{0}^{π / 4} e^{- x} {(\tan x)}^{50} d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x}$

$= \frac{50 \int_{0}^{π / 4} e^{- x} (\tan^{51} x + \tan^{49} x) d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x} = 50$

$∴$ $\frac{e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x} = 50$

Want Us to Email you About Special Offers & Updates?