JEE Advanced 2019 Paper 1
Exam JEE Advanced
Year 2019
Paper Paper 1
Subject Mathematics
Chapter Definite Integration JEE Advanced 2019 · Paper 1 · Mathematics · Definite Integration
If I = (2)/(π) ∫ _ - (π)/(4) ^ (π)/(4) dx (1 + e ^ x )(2 - 2x) then...
If
I = \dfrac{2}{\pi}\mathop{∫}_{- \dfrac{\pi}{4}}^{\dfrac{\pi}{4}}\dfrac{dx}{\left(1 + {e}^{\text{ \sin x }}\right)\left(2 - \cos 2x\right)} then find
27 I 2 27{I}^{2} 27 I 2 equals____
Step-by-step solution
I = 2 π ∫ − π 4 π 4 1 ( 1 + e sin x ) ( 2 − cos 2 x ) ⋅ d x I = \dfrac{2}{\pi}\mathop{∫}_{- \dfrac{\pi}{4}}^{\dfrac{\pi}{4}}\dfrac{1}{\left(1 + {e}^{\sin x}\right)\left(2 - \cos 2x\right)} ⋅ dx I = π 2 ∫ − 4 π 4 π ( 1 + e s i n x ) ( 2 − cos 2 x ) 1 ⋅ d x Using
∫ − a a f ( x ) ⋅ d x = ∫ 0 a ( f ( x ) + f ( − x ) ) ⋅ d x \mathop{∫}_{- a}^{a}f\left(x\right) ⋅ dx = \mathop{∫}_{0}^{a}\left(f\left(x\right) + f\left(- x\right)\right) ⋅ dx ∫ − a a f ( x ) ⋅ d x = ∫ 0 a ( f ( x ) + f ( − x ) ) ⋅ d x I = 2 π ∫ 0 π 4 ( 1 ( 1 + e sin x ) ( 2 − cos 2 x ) + 1 ( 1 + e − sin x ) ( 2 − cos 2 x ) ) ⋅ d x I = \dfrac{2}{\pi}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\left(\dfrac{1}{\left(1 + {e}^{\sin x}\right)\left(2 - \cos 2x\right)} + \dfrac{1}{\left(1 + {e}^{- \sin x}\right)\left(2 - \cos 2x\right)}\right) ⋅ dx I = π 2 ∫ 0 4 π ( ( 1 + e s i n x ) ( 2 − cos 2 x ) 1 + ( 1 + e − s i n x ) ( 2 − cos 2 x ) 1 ) ⋅ d x I = 2 π ∫ 0 π 4 ( 1 ( 1 + e sin x ) ( 2 − cos 2 x ) + e sin x ( e sin x + 1 ) ( 2 − cos 2 x ) ) ⋅ d x I = \dfrac{2}{\pi}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\left(\dfrac{1}{\left(1 + {e}^{\sin x}\right)\left(2 - \cos 2x\right)} + \dfrac{{e}^{\sin x}}{\left({e}^{\sin x} + 1\right)\left(2 - \cos 2x\right)}\right) ⋅ dx I = π 2 ∫ 0 4 π ( ( 1 + e s i n x ) ( 2 − cos 2 x ) 1 + ( e s i n x + 1 ) ( 2 − cos 2 x ) e s i n x ) ⋅ d x I = 2 π ∫ 0 π 4 d x 2 − cos 2 x I = \dfrac{2}{\pi}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{dx}{2 - \cos 2x} I = π 2 ∫ 0 4 π 2 − cos 2 x d x I = 2 π ∫ 0 π 4 d x 2 − 1 − tan 2 x 1 + tan 2 x = 2 π ∫ 0 π 4 ( 1 + tan 2 x ) d x 1 + 3 tan 2 x I = \dfrac{2}{\pi}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{dx}{2 - \dfrac{1 - \tan^{2}x}{1 + \tan^{2}x}} = \dfrac{2}{\pi}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{\left(1 + \tan^{2}x\right)dx}{1 + 3\tan^{2}x} I = π 2 ∫ 0 4 π 2 − 1 + tan 2 x 1 − tan 2 x d x = π 2 ∫ 0 4 π 1 + 3 tan 2 x ( 1 + tan 2 x ) d x I = 2 π ∫ 0 π 4 sec 2 x d x 1 + 3 t a n 2 x I = \dfrac{2}{\pi}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{\sec^{2}xdx}{1 + 3ta{n}^{2}x} I = π 2 ∫ 0 4 π 1 + 3 t a n 2 x sec 2 x d x Put
tan x = t \tan x = t tan x = t sec 2 x d x = d t \sec^{2} xdx = dt sec 2 x d x = d t and when
x = 0 ⇒ t = 0 x = 0 ⇒ t = 0 x = 0 ⇒ t = 0 x = π 4 ⇒ t = 1 x = \dfrac{\pi}{4} ⇒ t = 1 x = 4 π ⇒ t = 1 I = 2 π ∫ 0 1 d t 1 + 3 t 2 I = \dfrac{2}{\pi}\mathop{∫}_{0}^{1}\dfrac{dt}{1 + 3{t}^{2}} I = π 2 ∫ 0 1 1 + 3 t 2 d t I = 2 π ∫ 0 1 d t 1 + ( 3 t ) 2 I = \dfrac{2}{\pi}\mathop{∫}_{0}^{1}\dfrac{dt}{1 + {\left(\sqrt{3}t\right)}^{2}} I = π 2 ∫ 0 1 1 + ( 3 t ) 2 d t I = 2 π 1 3 [ t a n − 1 3 t ] 0 1 ( ∵ ∫ 0 1 d x a 2 + x 2 = 1 a tan − 1 x a + C ) I = \dfrac{2}{\pi}\dfrac{1}{\sqrt{3}}{\left[ta{n}^{- 1}\sqrt{3}t\right]}_{0}^{1}\left(∵ \mathop{∫}_{0}^{1}\dfrac{dx}{{a}^{2} + {x}^{2}} = \dfrac{1}{a}\tan^{- 1}\dfrac{x}{a} + C\right) I = π 2 3 1 [ t a n − 1 3 t ] 0 1 ( ∵ ∫ 0 1 a 2 + x 2 d x = a 1 tan − 1 a x + C ) I = 2 3 π [ t a n − 1 ( 3 ) − t a n − 1 ( 0 ) ] I = \dfrac{2}{\sqrt{3}\pi}\left[ta{n}^{- 1}\left(\sqrt{3}\right) - ta{n}^{- 1}\left(0\right)\right] I = 3 π 2 [ t a n − 1 ( 3 ) − t a n − 1 ( 0 ) ] I = 2 3 π × π 3 I = \dfrac{2}{\sqrt{3}\pi} \times \dfrac{\pi}{3} I = 3 π 2 × 3 π I = 2 3 3 I = \dfrac{2}{3\sqrt{3}} I = 3 3 2 27 I 2 = 27 × 4 27 = 4 27{I}^{2} = 27 \times \dfrac{4}{27} = 4 27 I 2 = 27 × 27 4 = 4
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