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CBSE Class 12-science Answered

Integrate    integral cos to the power of negative 1 end exponent open parentheses straight x close parentheses straight e to the power of straight x dx
Asked by manish.kkd4 | 19 Dec, 2018, 05:45: AM
answered-by-expert Expert Answer
Integration of this function excos-1x w.r.t. x  does not have any close form solution.
 
we have to expand exponetial function into series and integrate each term as shown below
begin mathsize 12px style integral e to the power of x cos to the power of negative 1 end exponent x space d x space equals integral cos to the power of negative 1 end exponent x open curly brackets 1 plus x plus fraction numerator x squared over denominator 2 factorial end fraction plus fraction numerator x cubed over denominator 3 factorial end fraction plus....... close curly brackets d x equals space integral cos to the power of negative 1 end exponent x space d x space plus space integral x space cos to the power of negative 1 end exponent x space d x space plus space 1 half integral x squared space cos to the power of negative 1 end exponent x space d x space plus space 1 over 6 integral x cubed cos to the power of negative 1 end exponent x space d x space plus space....... space t o space infinity i n space t h e space a b o v e space s e r i e s space w e space c a n space s u b s t i t u t e space t h e space f o l l o w i n g space i n t e g r a l s space f o r space e a c h space t e r m integral cos to the power of negative 1 end exponent x space d x space equals space x space cos to the power of negative 1 end exponent left parenthesis x right parenthesis space minus space square root of open parentheses 1 minus x squared close parentheses end root space plus space C integral x space cos to the power of negative 1 end exponent x space d x space equals space fraction numerator x squared cos to the power of negative 1 end exponent x over denominator 2 end fraction minus fraction numerator cos to the power of negative 1 end exponent x over denominator 4 end fraction minus fraction numerator x square root of 1 minus x squared end root over denominator 4 end fraction plus C integral x squared space cos to the power of negative 1 end exponent x space d x space equals space fraction numerator x cubed cos to the power of negative 1 end exponent x over denominator 3 end fraction minus fraction numerator open parentheses x squared plus 2 close parentheses square root of 1 minus x squared end root over denominator 9 end fraction plus C integral x to the power of m cos to the power of negative 1 end exponent x space d x space equals space fraction numerator x to the power of m plus 1 end exponent cos to the power of negative 1 end exponent x over denominator m plus 1 end fraction plus fraction numerator 1 over denominator m plus 1 end fraction integral fraction numerator x to the power of m plus 1 end exponent over denominator square root of 1 minus x squared end root end fraction d x space plus C end style
 
hence this integaral can be solved if |x| < 1 so that the series will be terminated after certain terms becomes negligibly small
Answered by Thiyagarajan K | 27 Dec, 2018, 12:23: PM

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