Sir / Mam 
Please suggest me another way to evaluate the Integral dx/(cos(x-a)cos(x-b)) other than the one suggested in the NCERT answers because I feel that it is quiet complicated. So please give any other alternative method and the logic behind multiplying and deviding the integrand in the question with sin(a-b). I mean that I want the explaination to understand the idea with what to multiply and devide the Integrand in such problems. I am totally struck up with this. Please Help me......

Asked by damarlasumanjali | 27th Jun, 2015, 10:21: AM

Expert Answer:

C o n s i d e r space t h e space g i v e n space i n t e g r a l comma I equals integral fraction numerator d x over denominator cos open parentheses x minus a close parentheses cos open parentheses x minus b close parentheses end fraction equals fraction numerator 1 over denominator sin open parentheses a minus b close parentheses end fraction integral fraction numerator sin open parentheses a minus b close parentheses d x over denominator cos open parentheses x minus a close parentheses cos open parentheses x minus b close parentheses end fraction equals fraction numerator 1 over denominator sin open parentheses a minus b close parentheses end fraction integral fraction numerator sin open square brackets open parentheses x minus b close parentheses minus open parentheses x minus a close parentheses close square brackets d x over denominator cos open parentheses x minus a close parentheses cos open parentheses x minus b close parentheses end fraction space space space space space space space space space open square brackets because sin open parentheses a minus b close parentheses equals sin open square brackets open parentheses x minus b close parentheses minus open parentheses x minus a close parentheses close square brackets close square brackets equals fraction numerator 1 over denominator sin open parentheses a minus b close parentheses end fraction open square brackets integral t a n open parentheses x minus b close parentheses d x minus integral t a n open parentheses x minus a close parentheses d x close square brackets space space space space space space open square brackets because sin open parentheses A minus B close parentheses equals sin A cos B minus cos A sin B close square brackets equals fraction numerator 1 over denominator sin open parentheses a minus b close parentheses end fraction open square brackets ln space s e c open parentheses x minus b close parentheses minus ln space s e c open parentheses x minus a close parentheses close square brackets plus C equals fraction numerator 1 over denominator sin open parentheses a minus b close parentheses end fraction open square brackets ln fraction numerator space s e c open parentheses x minus b close parentheses over denominator space s e c open parentheses x minus a close parentheses end fraction close square brackets plus C equals fraction numerator 1 over denominator sin open parentheses a minus b close parentheses end fraction open square brackets ln fraction numerator cos open parentheses x minus a close parentheses over denominator space cos open parentheses x minus b close parentheses end fraction close square brackets plus C
 
You please specify the page number in NCERT book.

Answered by Vimala Ramamurthy | 29th Jun, 2015, 10:53: AM

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