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Kindly consider the following

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Long answer type Questions as classified in NCERT Exemplar

    Sol:

    L e t             I = ∫ − π 4 π 4 l o g | s i n x + c o s x |   d x                                                                                                                             … ( i )                                     = ∫ − π 4 π 4 l o g | s i n ( π 4 − π 4 − x ) + c o s ( π 4 − π 4 − x ) |   d x             [ Using  ∫abf(x)dx=∫abf(a+b−x)dx ]                                       = ∫ − π 4 π 4 l o g | s i n ( − x ) + c o s x |   d x                                       = ∫ − π 4 π 4 l o g | c o s x − s i n x |   d x                                                                                                                         … ( i i ) A d d i n g     ( i )     a n d     ( i i )                           2 I = ∫ − π 4 π 4 l o g | c o s x + s i n x |   d x + ∫ − π 4 π 4 l o g | c o s x − s i n x |   d x                           2 I = ∫ − π 4 π 4 l o g | ( c o s x + s i n x ) ( c o s x − s i n x ) |   d x                           2 I = ∫ − π 4 π 4 l o g | c o s 2 x − s i n 2 x |   d x ∴                     2 I = ∫ − π 4 π 4 l o g c o s 2 x   d x                           2 I = 2 ∫ 0 π 4 l o g c o s 2 x   d x                             [ ? ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x     i f     f ( − x ) = f ( x ) ] ∴                           I = π ∫ 0 π 4 l o g c o s 2 x   d x P u t           2 x = t         d x = d t 2 W h e n     x = 0         ∴ t = 0 ;         w h e n     x = π 4         ∴ t = π 2                               I = 1 2 ∫ 0 π 2 l o g c o s t   d t                                     &thi

    O n     a d d i n g     ( i i i )     a n d     ( i v ) ,     w e     g e t                           2 I = 1 2 ∫ 0 π 2 ( l o g c o s t + l o g s i n t )   d t                             2 I = 1 2 ∫ 0 π 2 l o g s i n t c o s t   d t                             2 I = 1 2 ∫ 0 π 2 l o g 2 s i n t c o s t   d t 2                             2 I = 1 2 ∫ 0 π 2 ( l o g s i n 2 t − l o g 2 )   d t                             4 I = ∫ 0 π 2 l o g s i n 2 t   d t − ∫ 0 π 2 l o g 2   d t P u t             2 t = u         ⇒ 2 d t = d u         ⇒ d t = d u 2 ∴                       4 I = 1 2 ∫ 0 π l o g s i n u   d u − ∫ 0 π 2 l o g 2   d t             [ Changing  the  limit ]                             4 I = 1 2 × 2 ∫ 0 π 2 l o g s i n u   d u − l o g 2 [ t ] 0 π 2                               4 I = ∫ 0 π 2 l o g s i n u   d u − l o g 2 . π 2                               4 I = 2 I − π 2 . l o g 2                                                              

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