Question

Evaluate:
$integral subscript straight pi over 2 end subscript superscript straight pi space straight e to the power of straight x open parentheses fraction numerator 1 minus sinx over denominator 1 minus cosx end fraction close parentheses dx$

Solution

Let I = $integral subscript straight pi over 2 end subscript superscript straight pi space straight e to the power of straight x open parentheses fraction numerator 1 minus sinx over denominator 1 minus cosx end fraction close parentheses dx space equals space integral subscript straight pi over 2 end subscript superscript straight pi straight e to the power of straight x space open square brackets fraction numerator 1 minus 2 sin begin display style straight x over 2 end style cos begin display style straight x over 2 end style over denominator 2 sin squared begin display style straight x over 2 end style end fraction close square brackets dx$
$equals space integral subscript straight pi over 2 end subscript superscript straight pi space straight e to the power of straight x open square brackets fraction numerator 1 over denominator 2 sin squared begin display style straight x over 2 end style end fraction minus fraction numerator 2 sin begin display style straight x over 2 end style cos begin display style straight x over 2 end style over denominator 2 sin squared begin display style straight x over 2 end style end fraction close square brackets space dx space equals space integral subscript straight pi over 2 end subscript superscript straight pi straight e to the power of straight x open square brackets 1 half cosec squared straight x over 2 minus cot straight x over 2 close square brackets dx$
$equals space integral subscript straight pi over 2 end subscript superscript straight pi over 2 end superscript straight e to the power of straight x space open square brackets negative cot straight x over 2 plus 1 half cosec squared straight x over 2 close square brackets dx$
$equals space open square brackets straight e to the power of straight x space open parentheses negative cot straight x over 2 close parentheses close square brackets subscript straight pi over 2 end subscript superscript straight pi$ $open square brackets because space space space integral space straight e to the power of straight x space left curly bracket straight f left parenthesis straight x right parenthesis plus straight f apostrophe left parenthesis straight x right parenthesis right curly bracket dx space equals space straight e to the power of straight x space straight f left parenthesis straight x right parenthesis close square brackets$
$equals negative open square brackets straight e to the power of straight x open parentheses negative cot straight x over 2 close parentheses close square brackets subscript straight pi over 2 end subscript superscript straight pi space equals space minus open square brackets straight e to the power of straight pi cot straight pi over 2 minus straight e to the power of straight pi over 2 end exponent cot straight pi over 4 close square brackets equals negative open square brackets straight e to the power of straight pi space cross times 0 minus straight e to the power of straight pi over 2 end exponent cross times 1 close square brackets space equals space straight e to the power of straight pi over 2 end exponent$

For Daily Free Study Material Join wiredfaculty

Integrals

Evaluate:
$integral subscript straight pi over 2 end subscript superscript straight pi space straight e to the power of straight x open parentheses fraction numerator 1 minus sinx over denominator 1 minus cosx end fraction close parentheses dx$

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction