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∫ − π 2 π 2 8 2 c o s x ( 1 + e s i n x ) ( 1 + s i n 4 x ) dx = ap + b log ( 3 + 2 2 )  

Option 1 -

4

Option 2 -

6

Option 3 -

8

Option 4 -

2

0 2 Views | Posted a week ago
Asked by Shiksha User

  • 1 Answer

  • Correct Option - 1


    Detailed Solution:

    ∫ − π 2 π 2 8 2 c o s x ( 1 + e s i n x ) ( 1 + s i n 4 x ) dx

    = ∫ 0 π 2 { ( 8 2 c o s x ( 1 + e s i n x ) ( 1 + s i n 4 x ) + 8 2 c o s x ( 1 + e − s i n x ) ( 1 + s i n 4 x ) ) } d x            

    = 8 2 ∫ 0 π 2 c o s x 1 + s i n 4 x d x            

    Let sin x = t

    I = 8 2 ∫ 0 1 d t 1 + t 4            

    = 4 2 ∫ 0 1 ( 1 + 1 t 2 ) − ( 1 − 1 t 2 ) t 2 + 1 t 2 d t       

    = 4 2 ∫ 0 1 ( 1 + 1 t 2 ) d t ( t − 1 t ) 2 + 2 − 4 2 ∫ 0 1 ( 1 − 1 t 2 ) d t ( t + 1 t ) 2 − 2            

    = 4 2 ⋅ 1 2 ( t a n − 1 t − 1 t 2 ) 0 1 − 4 2 ⋅ 1 2 2 [ l o g | t + 1 t − 2 t + 1 t + 2 | ] 0 1         

    = 2 π − 2 l o g | 2 − 2 2 + 2 |        

    = 2 π + 2 l o g ( 3 + 2 2 )           

    a = b = 2

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The value of the integral

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