Exams Joint Entrance Examination Mathematics Definite Integration Question 33 Question 33 of 246 Medium MCQ
If ∫ π 2 − π 2 96 x 2 cos 2 x ( 1 + e x ) d x = π ( α π 2 + β ) , α , β ∈ Z {\int }_{\frac{\pi }{2}}^{-\frac{\pi }{2}}\frac{96{x}^{2}{\cos }^{2}x}{(1+{e}^{x})}dx=\pi (\alpha {\pi }^{2}+\beta ),\alpha ,\beta \in Z ∫ 2 π − 2 π ( 1 + e x ) 96 x 2 c o s 2 x d x = π ( α π 2 + β ) , α , β ∈ Z , then ( α + β ) 2 (\alpha +\beta {)}^{2} ( α + β ) 2 equals
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∫ − π 2 π 2 96 x 2 cos 2 x ( 1 + e x ) d x \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{96{x}^{2}{\cos }^{2}x}{(1+{e}^{x})}dx\, ∫ − 2 π 2 π ( 1 + e x ) 96 x 2 c o s 2 x d x (Apply King Property)
∫ π 2 0 96 x 2 cos 2 x = 48 ∫ π 2 0 x 2 ( 1 + cos 2 x ) d x 48 [ ( x 3 3 ) π / 2 0 + ∫ π 2 0 x 2 1 cos I I 2 x d x ] ⇒ π ( 2 π 2 − 12 ) ⇒ α = 2 , β = − 12 ⇒ ( α + β ) 2 = 100 \begin{gathered} {\int }_{\frac{\pi }{2}}^{0}96{x}^{2}{\cos }^{2}x=48{\int }_{\frac{\pi }{2}}^{0}{x}^{2}(1+\cos 2x)dx \\ 48[{(\frac{{x}^{3}}{3})}_{\pi /2}^{0}+{\int }_{\frac{\pi }{2}}^{0}{x}_{2}^{1}{\cos }_{II}2xdx] \\ \Rightarrow \pi (2{\pi }^{2}-12) \\ \Rightarrow \alpha =2,\beta =-12 \\ \Rightarrow (\alpha +\beta {)}^{2}=100 \end{gathered} ∫ 2 π 0 96 x 2 cos 2 x = 48 ∫ 2 π 0 x 2 ( 1 + cos 2 x ) d x 48 [ ( 3 x 3 ) π /2 0 + ∫ 2 π 0 x 2 1 cos I I 2 x d x ] ⇒ π ( 2 π 2 − 12 ) ⇒ α = 2 , β = − 12 ⇒ ( α + β ) 2 = 100 Answer the question to unlock