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Integrals

Question
CBSEENMA12036070

If the integral integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x space equals space straight x
space plus space straight a space log space vertical line space sin space straight x minus space 2 space cos space straight x vertical line space plus space straight k comma  then an equal to 

  • -1

  • -2

  • 1

  • 2

Solution

D.

2

integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x space equals space straight x space plus space straight a space log space vertical line space sin space straight x minus space 2 space cos space straight x vertical line space plus space straight k comma space... space left parenthesis straight i right parenthesis
Now, let us assume that I,
straight I space equals integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x
Multiplying by cos x in numerator and denominator, we get
straight I space equals space integral fraction numerator 5 space sin space straight x over denominator sin space straight x minus 2 space cos space straight x end fraction straight d space straight x
This special integration requires special substitution of type
N' = A (D') +straight B space open parentheses fraction numerator dD apostrophe over denominator dx end fraction close parentheses
⇒ Let 5 sin x = (A + 2B) sin x + (B- 2A) cos x
Comapring the coefficients of sin x and cosx, 
we get
A +2B =5 and B- 2A = 0
Solving the above two equations in A and B
we get
A = 1 an B= 2
⇒ 5 sin x = ( sin x - 2 cos x)+ 2 (cos x + 2 sin x)
rightwards double arrow space straight I space equals space integral fraction numerator 5 space sin space straight x over denominator sin space straight x minus space 2 space cos space straight x end fraction dx
space equals space integral fraction numerator sin space straight x minus space 2 space cos space straight x right parenthesis space plus space 2 space left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis space dx over denominator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis end fraction
rightwards double arrow space straight I space equals space integral fraction numerator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis plus space 2 space left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis over denominator sin space straight x minus space 2 cos space straight x end fraction space dx
rightwards double arrow space straight I space equals space integral fraction numerator sin space straight x minus space 2 space cos space straight x over denominator sin space straight x minus space 2 space cos space straight x end fraction dx space plus space 2 space integral fraction numerator left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis over denominator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis end fraction dx
rightwards double arrow space straight I space equals integral space 1 space dx space plus space 2 space integral fraction numerator straight d left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis over denominator left parenthesis sin space straight x minus space 2 space cos space straight x right parenthesis space end fraction
rightwards double arrow space straight I space space equals space straight x space plus 2 space log space vertical line sin space straight x minus space 2 space cos space straight x right parenthesis vertical line space plus straight k space... space left parenthesis ii right parenthesis
where k is the constant of integration. Now, by comparing the value of l in eq. (i) and (ii) we get a = 2

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