CBSE Class 12-science Answered
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![question image](http://images.topperlearning.com/topper/new-ate/top_mob16967533231100895497e33c03e-1ec3-43a7-a298-ebd01cf70d6d.jpg)
Asked by tanuspatel5 | 08 Oct, 2023, 13:52: PM
![begin mathsize 14px style L e t space I space equals integral fraction numerator s e c x space space d x over denominator square root of cos left parenthesis 2 x plus alpha right parenthesis plus cos alpha end root end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/e498ee190b73c57ad28e61f5fffc5460.png)
we have
![begin mathsize 14px style cos left parenthesis 2 x plus alpha space right parenthesis space equals space cos left parenthesis 2 x right parenthesis space cos alpha space minus space sin left parenthesis 2 x right parenthesis space sin alpha end style](https://images.topperlearning.com/topper/tinymce/cache/0328fd9f9417941cedeac10a13b5c547.png)
![begin mathsize 14px style cos left parenthesis 2 x right parenthesis space equals space fraction numerator 1 space minus space tan squared x over denominator 1 plus tan squared x end fraction space space space semicolon space space space space space sin left parenthesis 2 x right parenthesis space equals space fraction numerator 2 space tan left parenthesis x right parenthesis over denominator 1 plus tan squared x end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/010dd4881c86e7f40a907f31e9f5b858.png)
Using the above formulas , integration can be written as
![begin mathsize 14px style I space equals space integral fraction numerator s e c squared x space d x over denominator square root of left parenthesis 1 minus tan squared x right parenthesis space cos alpha space minus space left parenthesis space 2 space tan x space right parenthesis space sin alpha space plus space left parenthesis 1 plus tan squared x right parenthesis space cos alpha end root end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/3c0822644c471c9ba66d4d941e37becb.png)
After simplification, above integration is written as
![begin mathsize 14px style I space equals space integral fraction numerator s e c squared x space d x over denominator square root of 2 space cos alpha space minus space left parenthesis 2 sin alpha right parenthesis tan x end root end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/1b230f45d791e0cb5c41ddce745ebf17.png)
Let u = tan(x) , then du = sec2x dx
Let ( 2 cosα ) = a and ( 2 sinα ) = b
Using these substitutions , we get
![begin mathsize 14px style I space equals space integral fraction numerator d u over denominator square root of a space minus space b space u end root end fraction space equals space minus 2 over b space square root of a space minus space b space u end root space plus space C end style](https://images.topperlearning.com/topper/tinymce/cache/a07f1c1940b5426ec3255b94ae0bf3a3.png)
where C is constant of integration .
![begin mathsize 14px style I space equals space minus fraction numerator 1 over denominator sin alpha end fraction space square root of left parenthesis 2 space cos alpha right parenthesis space minus space left parenthesis 2 space sin alpha right parenthesis space tan x end root space plus space C end style](https://images.topperlearning.com/topper/tinymce/cache/28d3c872cdb4b6dd5e60b460c6e12555.png)
Answered by Thiyagarajan K | 09 Oct, 2023, 15:41: PM
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