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integrate 2sinx-cosx+3/sinx-2cosx+1 dx
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Asked by ayush9828450465 | 15 Oct, 2020, 10:17: AM
answered-by-expert Expert Answer
Given integration :- begin mathsize 14px style I space equals space integral fraction numerator 2 space sin x space minus space cos x space plus 3 over denominator sin x space minus space 2 space cos x space plus 1 end fraction d x end style
Let u = sinx - 2cosx +1 ,  then du = ( cosx+2sinx ) dx
 
numerator of integrand :-  (2 sinx - cosx +3 ) dx = ( 2sinx + cosx -2cosx +3 ) dx =  du + (3 - 2 cosx) dx
 
Hence given intgration becomes
 
begin mathsize 14px style I space equals space integral fraction numerator d u over denominator u end fraction plus integral fraction numerator left parenthesis space 3 space minus space 2 space cos x right parenthesis space d x over denominator left parenthesis space sin x space minus space 2 space cos x space plus 1 space right parenthesis end fraction space equals space log left parenthesis space sin x space minus space 2 space cos x space plus space 1 space right parenthesis space plus space I subscript 1 end style .....................(1)
Where 
 
begin mathsize 14px style I subscript 1 space equals space integral fraction numerator left parenthesis space 3 space minus space 2 space cos x right parenthesis space d x over denominator left parenthesis space sin x space minus space 2 space cos x space plus 1 space right parenthesis end fraction space end style......................(2)
To integrate above, we use the substitutions ,  t = tan(x/2) ,  sinx = 2t / ( 1+t2 ) , cosx = (1-t2 )/(1+t2 ) , dx = 2dt / ( 1+ t2 )
 
By using the above substitutions, we get
 
begin mathsize 14px style I subscript 1 space equals space 2 integral fraction numerator left parenthesis space 1 space plus space 5 space t squared right parenthesis space space d t over denominator left parenthesis space 3 t squared space plus space 2 space t space minus 1 space right parenthesis space left parenthesis space 1 space plus space t squared right parenthesis end fraction space end style.........................(3)
Integrand in the above integration is written as partial fraction as given below
 
begin mathsize 14px style fraction numerator left parenthesis space 1 space plus space 5 space t squared space right parenthesis over denominator left parenthesis space 3 t squared plus 2 t space minus 1 space right parenthesis left parenthesis 1 plus t squared right parenthesis end fraction space equals space fraction numerator left parenthesis space 1 space plus space 5 t squared space right parenthesis over denominator left parenthesis space 1 space plus space t space right parenthesis space left parenthesis space 3 space t space minus space 1 space right parenthesis space left parenthesis space 1 space plus space t squared right parenthesis end fraction space equals space fraction numerator A over denominator 1 plus t end fraction space plus space fraction numerator B over denominator 3 t space minus 1 end fraction space plus space fraction numerator C space t space plus D over denominator 1 space plus space t squared end fraction end style  ...................(4)
 
Hence we write
 
(1+5 t2 ) = A ( 3t -1 ) ( 1 + t2 ) + B ( 1+t) (1 +t2 ) + ( C t + D ) ( 1 + t ) ( 3 t -1 )  ........................(5)
 
if we put t = 1/3 in above eqn.(5) , we get B = 21/20
 
if we put t = -1 in above eqn.(5) , we get A = -3/4 
 
if we put t = 0 in above eqn.(5) , we get  1 = -A  +B -D  = (3/4) + 21/20 -D , hence D = 4/5
 
if we put t = 1 in above eqn.(5) , we get  6 = 4A  +4B  +( C+D)4  , By substituting values for A , B and D , we get C = 2/5
 
using the values A, B, C and D , we rewrite integration as given in eqn.(3) as
 
begin mathsize 14px style I subscript 1 space equals space minus 3 over 2 integral fraction numerator d t over denominator 1 plus t end fraction space plus space 7 over 10 integral fraction numerator 3 space d t over denominator 3 t space minus 1 end fraction space plus 2 over 5 integral fraction numerator 2 t d t over denominator 1 plus t squared end fraction space plus space 8 over 5 integral fraction numerator d t over denominator 1 plus t squared end fraction end style
Integration of each term in above equation is known . 
Hence after integration of each term , by using substitution t = tan(x/2) and after simplification , we get
 
begin mathsize 14px style I subscript 1 space equals space log fraction numerator left parenthesis space 3 space tan left parenthesis x divided by 2 right parenthesis space minus 1 space right parenthesis to the power of 0.7 end exponent s e c to the power of 0.8 end exponent left parenthesis x divided by 2 right parenthesis over denominator left parenthesis space 1 space plus space tan left parenthesis x divided by 2 right parenthesis right parenthesis to the power of 1.5 end exponent end fraction space plus space 4 over 5 x end style
 
Hence using eqn.(1) , given integration becomes 
 
 
 begin mathsize 14px style I space equals space log space left parenthesis space sin x space minus space 2 space cos x space plus 1 space right parenthesis space plus space log fraction numerator left parenthesis space 3 space tan left parenthesis x divided by 2 right parenthesis space minus 1 space right parenthesis to the power of 0.7 end exponent s e c to the power of 0.8 end exponent left parenthesis x divided by 2 right parenthesis over denominator left parenthesis space 1 space plus space tan left parenthesis x divided by 2 right parenthesis right parenthesis to the power of 1.5 end exponent end fraction space plus space 4 over 5 x end style
 
begin mathsize 14px style I space equals space log fraction numerator left parenthesis space sin x space minus space 2 space cos x space plus 1 space right parenthesis space left parenthesis space 3 space tan left parenthesis x divided by 2 right parenthesis space minus 1 space right parenthesis to the power of 0.7 end exponent space s e c to the power of 0.8 end exponent left parenthesis x divided by 2 right parenthesis over denominator left parenthesis space 1 space plus space tan left parenthesis x divided by 2 right parenthesis right parenthesis to the power of 1.5 end exponent end fraction space plus space 4 over 5 x end style
 
Answered by Thiyagarajan K | 15 Oct, 2020, 04:00: PM

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