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CBSE Class 12-science Answered

SIR I HAVE ATTACHED THE QUESTION BELOW PLEASE HAVE A LOOK ON IT. 
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Asked by PARTHAGRAWAL914 | 14 Dec, 2018, 04:05: PM
answered-by-expert Expert Answer
begin mathsize 12px style I space equals space integral fraction numerator x squared plus cos squared x over denominator 1 plus x squared end fraction space cos e c squared x space d x space equals space integral fraction numerator x squared cos e c squared x over denominator 1 plus x squared end fraction space d x space plus space integral fraction numerator c o t squared x over denominator 1 plus x squared end fraction space d x
I space equals space integral fraction numerator x squared over denominator 1 plus x squared end fraction space cos e c squared x space d x space plus space integral fraction numerator c o t squared x over denominator 1 plus x squared end fraction space d x space
I space equals space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus x squared end fraction close parentheses cos e c squared x space d x space plus space integral fraction numerator c o t squared x over denominator 1 plus x squared end fraction space d x
I space equals space integral cos e c squared x space d x space plus space integral fraction numerator open parentheses c o t squared x space minus space cos e c squared x close parentheses over denominator 1 plus x squared end fraction space d x
I space equals space integral fraction numerator d x over denominator begin display style sin squared x end style end fraction space minus space integral fraction numerator d x over denominator 1 plus x squared end fraction space space equals space I subscript 1 space minus space tan to the power of negative 1 end exponent x space plus C
w h e r e space I subscript 1 space equals space integral fraction numerator d x over denominator begin display style sin squared x end style end fraction space space semicolon space
t o space s o l v e space I subscript 1 space w e space u s e space s u b s t i t u t i o n space x space equals space tan to the power of negative 1 end exponent t space comma space h e n c e space d x space equals space fraction numerator d t over denominator 1 plus t squared end fraction space a n d space sin squared x space equals fraction numerator t squared over denominator 1 plus t squared end fraction
I subscript 1 space equals space integral fraction numerator d t over denominator t squared end fraction space space equals space minus fraction numerator 1 over denominator tan left parenthesis begin display style x right parenthesis end style end fraction plus C
I space equals space minus c o t left parenthesis x right parenthesis minus tan to the power of negative 1 end exponent x space plus space C end style
Answered by Thiyagarajan K | 15 Dec, 2018, 07:02: AM
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