CBSE Class 12-science Answered
integrate
![question image](http://images.topperlearning.com/topper/new-ate/top_mob16741329661643791873IMG_20230119_182453.jpg)
Asked by ritiksingh84835808 | 19 Jan, 2023, 18:26: PM
![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](https://images.topperlearning.com/topper/tinymce/cache/1ef6da5f5824ab002e35eef35a1e3f6f.png)
Let us use the substitution " t = tan(θ/2) "
we get ,
;
; ![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](https://images.topperlearning.com/topper/tinymce/cache/397ab0c797bb60261aab69bee9d15972.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/34625470d3eaa426292547231bf84647.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/6e9484d20d18691ac6dc6ab778a39c6a.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/397ab0c797bb60261aab69bee9d15972.png)
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](https://images.topperlearning.com/topper/tinymce/cache/96cbb03f283942eee40c586d1489aa2a.png)
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](https://images.topperlearning.com/topper/tinymce/cache/8c332f15b2d6afa64854c3c7b097426c.png)
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](https://images.topperlearning.com/topper/tinymce/cache/24ce458b95a1bed6037216b01e88a649.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/1dc5924495e1eb8b6268bccf86204ad2.png)
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](https://images.topperlearning.com/topper/tinymce/cache/635fc1cad3cb05748b5a5b487271ead7.png)
I = ![begin mathsize 14px style pi divided by 4 end style](https://images.topperlearning.com/topper/tinymce/cache/157d96effc7477f8baae6b978716b541.png)
![begin mathsize 14px style pi divided by 4 end style](https://images.topperlearning.com/topper/tinymce/cache/157d96effc7477f8baae6b978716b541.png)
Answered by Thiyagarajan K | 19 Jan, 2023, 20:45: PM
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