CBSE Class 12-science Answered

integrate

Asked by ritiksingh84835808 | 19 Jan, 2023, 06:26: PM

Expert Answer

begin mathsize 14px style I space equals space integral subscript o superscript pi divided by 2 end superscript fraction numerator cos theta over denominator cos theta plus sin theta end fraction d theta end style

Let us use the substitution " t = tan(θ/2) "

we get ,

begin mathsize 14px style cos theta space equals space fraction numerator 1 space minus space t squared over denominator 1 plus t squared end fraction end style

;

begin mathsize 14px style sin theta space equals space fraction numerator 2 space t over denominator 1 space plus space t squared end fraction end style

;

begin mathsize 14px style d theta space equals space fraction numerator 2 space d t over denominator 1 space plus space t squared end fraction end style

By using above substitutions, we get

begin mathsize 14px style I space equals space 2 space integral subscript o superscript 1 fraction numerator left parenthesis 1 minus t squared right parenthesis space d t over denominator left parenthesis 1 plus t squared right parenthesis space left parenthesis space 1 space minus space t squared plus 2 t space right parenthesis end fraction end style

Let us resolve the above integrand into partial fraction as shown below

begin mathsize 14px style fraction numerator left parenthesis 1 minus t squared right parenthesis space over denominator left parenthesis 1 plus t squared right parenthesis space left parenthesis space 1 space minus space t squared plus 2 t space right parenthesis end fraction space equals space fraction numerator A space t plus B over denominator 1 plus t squared end fraction plus fraction numerator C space t space plus D over denominator left parenthesis 1 space minus space t squared plus 2 t right parenthesis end fraction end style

we get A = -1/2 , B = 1/2 , C = -1/2 , D = 1/2

Then we get

begin mathsize 14px style I space equals space integral subscript o superscript 1 fraction numerator left parenthesis 1 minus t right parenthesis space d t over denominator left parenthesis 1 plus t squared right parenthesis space end fraction plus integral subscript o superscript 1 fraction numerator left parenthesis 1 minus t right parenthesis space d t over denominator space left parenthesis space 1 space minus space t squared plus 2 t space right parenthesis end fraction end style

begin mathsize 14px style I space equals space integral subscript o superscript 1 fraction numerator space d t over denominator left parenthesis 1 plus t squared right parenthesis space end fraction space minus space 1 half integral subscript o superscript 1 fraction numerator 2 space t space d t over denominator left parenthesis 1 plus t squared right parenthesis space end fraction plus 1 half integral subscript o superscript 1 fraction numerator left parenthesis 2 minus 2 t right parenthesis space d t over denominator space left parenthesis space 1 space minus space t squared plus 2 t space right parenthesis end fraction end style

we get from above integration ,

begin mathsize 14px style I space equals space open parentheses tan to the power of negative 1 end exponent t close parentheses subscript 0 superscript 1 space minus space 1 half open parentheses log left parenthesis 1 plus t squared right parenthesis close parentheses subscript o space end subscript superscript 1 plus 1 half open parentheses log left parenthesis 1 minus t squared plus 2 t right parenthesis close parentheses subscript o space end subscript superscript 1 end style

I =

begin mathsize 14px style pi divided by 4 end style

Answered by Thiyagarajan K | 19 Jan, 2023, 08:45: PM

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