If the integral 525 0 2 sin 2 x cos 11 2 x ( 1 + cos 5 2 x ) 1 2 d x is equal to ( n 2 - 64 ) , then n is equal to [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 the integral $525 \int_{0}^{\frac{π}{2}} \sin 2 x \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2} - 64),$ then $n$ is equal to ______ . [2024]

Ans.
(176)

Let $I = 525 \int_{0}^{\frac{π}{2}} \sin 2 x \cdot \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$

Put $\cos x = t^{2}$

$- \sin x d x = 2 t d t$

$I = - 525 \int_{1}^{0} 2 \sin x \cdot t^{2} \cdot t^{11} {(1 + t^{5})}^{\frac{1}{2}} \cdot \frac{2 t d t}{\sin x}$

$= 525 \times 4 \int_{0}^{1} t^{3} \cdot t^{11} {(1 + t^{5})}^{\frac{1}{2}} d t$

$I = 2100 \int_{0}^{1} t^{14} \sqrt{(1 + t^{5})} d t$

Let $1 + t^{5} = k^{2} \Rightarrow 5 t^{4} d t = 2 k d k \Rightarrow t^{4} d t = \frac{2 k d k}{5}$

$∴ I = 2100 \int_{1}^{\sqrt{2}} {(k^{2} - 1)}^{2} \cdot k \cdot \frac{2 k d k}{5}$

$= \frac{2100 \times 2}{5} \int_{1}^{\sqrt{2}} k^{2} {(k^{2} - 1)}^{2} d k$

   $= 420 \times 2 \int_{1}^{\sqrt{2}} (k^{6} - 2 k^{4} + k^{2}) d k$

   $= 840 {[\frac{k^{7}}{7} - \frac{2 k^{5}}{5} + \frac{k^{3}}{3}]}_{1}^{\sqrt{2}}$

$= 840 [\frac{8 \sqrt{2}}{7} - \frac{8 \sqrt{2}}{5} + \frac{2 \sqrt{2}}{3} - (\frac{1}{7} - \frac{2}{5} + \frac{1}{3})]$

   $= 840 [\frac{22 \sqrt{2} - 8}{105}] = 8 [22 \sqrt{2} - 8] = 176 \sqrt{2} - 64$

So, $n = 176$

Want Us to Email you About Special Offers & Updates?