CBSE Class 12-science Answered

Asked by tanuspatel5 | 08 Oct, 2023, 13:52: PM

Expert Answer

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

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

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

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

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

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

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

Answered by Thiyagarajan K | 09 Oct, 2023, 15:41: PM

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