Question

SOlVE THIS QUESTION....SIS AND...BRO..​

Answer 1

GIVEN :–

$\\ \implies\bf \: I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \frac{1}{(1 + {e}^{ \sin x } )(2 - \cos 2x)}.dx$

TO FIND :–

• Value of 27I² = ?

SOLUTION :–

$\\ \implies\bf \: I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \frac{1}{(1 + {e}^{ \sin x } )(2 - \cos 2x)}.dx \: \: \: - - - eq.(1)$

• Using property –

$\\ \implies\bf \int_{a}^{b}f(x).dx = \int_{a}^{b}f(a + b - x).dx$

• So that –

$\\ \implies\bf \: I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \frac{1}{(1 + {e}^{ - \sin x } )(2 - \cos 2x)}.dx$

$\\ \implies\bf \: I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{ {e}^{ \sin x} }{( {e}^{ \sin x } + 1)(2 - \cos 2x)} .dx\: \: \: - - - eq.(2)$

• Now add eq.(1) and eq.(2) –

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{ ({e}^{ \sin x} + 1)}{( {e}^{ \sin x } + 1)(2 - \cos 2x)} .dx$

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{1}{2 - \cos 2x}.dx$

• We know that –

$\\ \implies\bf \: \cos(2x) = 2\cos^{2} - 1$

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{1}{2 - \cos 2x}.dx$

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{1}{2 - ( 2 \cos^{2} x - 1)}.dx$

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{1}{2 - 2 \cos^{2} x +1}.dx$

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{1}{3 - 2 \cos^{2} x }.dx$

• We should write this as –

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{ { \sec}^{2}x }{3{ \sec}^{2}x- 2 }.dx$

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4}} \dfrac{ { \sec}^{2}x }{3(1 + { \tan}^{2}x)- 2 }.dx$

• Put tan(x) = t –

$\\ \implies\bf \: (\sec^{2} x).dx = dt$

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ -1}^{1} \dfrac{dt}{3(1 + {t}^{2})- 2 }$

$\\ \implies\bf \: 2I \: = \dfrac{2}{\pi} \int_{ -1}^{1} \dfrac{dt}{3 + 3{t}^{2}- 2 }$

$\\ \implies\bf \: I \: = \dfrac{1}{\pi} \int_{ -1}^{1} \dfrac{dt}{3{t}^{2} + 1 }$

$\\ \implies\bf \: I \: = \dfrac{1}{3\pi} \int_{ -1}^{1} \dfrac{dt}{{t}^{2} + { (\frac{1}{ \sqrt{3} } )}^{2} }$

• Using identity —

$\\ \implies\bf \int\dfrac{dt}{{t}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \left( \frac{t}{a} \right) + c$

$\\ \implies\bf \: I \: = \dfrac{\sqrt{3}}{3\pi} \left[ { \tan}^{ - 1}( \sqrt{3} t) \right]_{ -1}^{1}$

$\\ \implies\bf \: I \: = \dfrac{1}{\sqrt{3}\pi} \left[ { \tan}^{ - 1}( \sqrt{3} ) - { \tan}^{ - 1}( - \sqrt{3} ) \right]$

$\\ \implies\bf \: I \: = \dfrac{1}{\sqrt{3}\pi} \left[ { \tan}^{ - 1}( \sqrt{3} ) + { \tan}^{ - 1}( \sqrt{3} ) \right]$

$\\ \implies\bf \: I \: = \dfrac{1}{\sqrt{3}\pi} \left[2 { \tan}^{ - 1}( \sqrt{3} ) \right]$

$\\ \implies\bf \: I \: = \dfrac{1}{\sqrt{3}\pi} \left[2 \times \dfrac{\pi}{3} \right]$

$\\ \implies\bf \: I \: = \dfrac{2}{3\sqrt{3}}$

$\\ \implies\bf \: I ^{2} \: = \dfrac{4}{27}$

$\\ \implies \large{\boxed{\bf 27{I}^{2} \: =4}}$