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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \large\frac{1}{\sin (x-a) \cos (x-b)} $$ dx$

$(a)\;\frac{1}{\cos (a-b)}\bigg[\log |\sin (x-a)|+\log | \sec (x-b)|\bigg]+c \qquad (b)\;\frac{1}{\cos (a-b)}\bigg[\log |\sin (x-a)|-\log | \sec (x-b)|\bigg]+c \\ (c)\;\frac{1}{\cos (a+b)}\bigg[\log |\sin (x-a)|+\log | \sec (x-b)|\bigg]+c \qquad (d)\;\frac{1}{\sin (a-b)}\bigg[\log |\sin (x-a)|+\log | \sec (x-b)|\bigg ]+c $

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1 Answer

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$\int \large\frac{1}{\sin (x-a) \cos (x-b)} $$ dx$
=> $\large\frac{1}{\cos (a-b)} \int \large\frac{\cos (a-b)}{\sin (x-a) \cos (x-b)}$$dx$
Multiply $\cos (a-b)$ in numerator and denominator
=> $\large\frac{1}{\cos (a-b)} \int \large\frac{\cos [(x-b)-(x-a)]}{\sin (x-a) \cos (x-b)}$
=> $\large\frac{1}{\cos (a-b)} \int \bigg\{ \large\frac{\cos (x-b) \cos (x-b)+ \sin (x-b) \sin (x-a)}{\sin (x-a) \cos (x-b)}\bigg \}$
=> $\large\frac{1}{\cos (a-b)}$$ \int [\cot (x-a)+ \tan (x-b)]dx$
$\large\frac{1}{\cos (a-b)}$$\bigg[\log |\sin (x-a)|+\log | \sec (x-b)|\bigg]+c$
Hence a is the correct answer.
answered Dec 24, 2013 by meena.p
 
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