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sec²x/(secx+thanx)⁹/²

Asked by snehasharma796840 | 30 Oct, 2023, 10:24: PM

Expert Answer

Let sec(x) + tan(x) = t ................................(1)

By differentiating eqn.(1) , we get

sec(x) tan(x) dx + sec²x dx = dt

sec(x) [ sec(x) + tanx(x) ] dx= dt

begin mathsize 14px style s e c left parenthesis x right parenthesis space d x space equals space fraction numerator d t over denominator s e c left parenthesis x right parenthesis plus tan left parenthesis x right parenthesis end fraction space equals space fraction numerator d t over denominator t end fraction end style

we know that , sec²(x) - tan²(x) = 1

[ sec(x) + tan(x) ] [ sec(x) - tan(x) ] = 1

begin mathsize 14px style s e c left parenthesis x right parenthesis space minus space tan left parenthesis x right parenthesis space equals space fraction numerator 1 over denominator left square bracket space s e c left parenthesis x right parenthesis space plus space tan left parenthesis x right parenthesis space right square bracket end fraction space equals space 1 over t end style

...........................(2)

By adding eqn.(1) and eqn.(2) , we get ,

begin mathsize 14px style s e c left parenthesis x right parenthesis space equals space 1 half space open parentheses t space plus space 1 over t close parentheses end style

Hence the given integral is

begin mathsize 14px style I space equals space integral fraction numerator s e c squared left parenthesis x right parenthesis space d x over denominator open parentheses s e c left parenthesis x right parenthesis plus tan left parenthesis x right parenthesis close parentheses to the power of begin display style bevelled 9 over 2 end style end exponent end fraction space equals space integral fraction numerator s e c left parenthesis x right parenthesis space over denominator open parentheses s e c left parenthesis x right parenthesis plus tan left parenthesis x right parenthesis close parentheses to the power of bevelled 9 over 2 end exponent end fraction space s e c left parenthesis x right parenthesis d x end style

begin mathsize 14px style I space equals space 1 half integral fraction numerator open parentheses t plus begin display style 1 over t end style close parentheses over denominator t to the power of begin display style bevelled 9 over 2 end style end exponent end fraction space fraction numerator d t over denominator t end fraction space equals space 1 half integral open parentheses t to the power of bevelled fraction numerator negative 9 over denominator 2 end fraction end exponent space plus space t to the power of bevelled fraction numerator negative 13 over denominator 2 end fraction end exponent close parentheses space d t end style

begin mathsize 14px style I space equals space 1 half open square brackets open parentheses fraction numerator negative 2 over denominator 7 end fraction close parentheses space t to the power of bevelled fraction numerator negative 7 over denominator 2 end fraction end exponent space plus space open parentheses fraction numerator negative 2 over denominator 11 end fraction close parentheses space t to the power of bevelled fraction numerator negative 11 over denominator 2 end fraction end exponent close square brackets space plus space C end style

You can substitute for t in above expression and simplify if possible to get final answer

Answered by Thiyagarajan K | 31 Oct, 2023, 05:11: PM

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