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CBSE Class 12-science Answered

plzz solve this
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Asked by tyagisrishti56 | 12 Oct, 2020, 09:36: PM
answered-by-expert Expert Answer
begin mathsize 14px style I space equals space integral fraction numerator x squared left parenthesis space x space s e c squared x space plus space tan x space right parenthesis over denominator open parentheses x space tan x space plus 1 close parentheses squared end fraction d x end style
Let u = ( x tanx +1 )  ,   du = ( x sec2x + tanx ) dx
 
Then we have
 
begin mathsize 14px style fraction numerator left parenthesis space x space s e c squared x space plus space tan x space right parenthesis space d x over denominator left parenthesis x space tan x space plus space 1 right parenthesis squared end fraction space equals space fraction numerator d u over denominator u squared end fraction space equals space minus d left parenthesis space u to the power of negative 1 to the power of space end exponent right parenthesis end style
Then integration becomes
 
begin mathsize 14px style I space equals space minus integral x squared space d open parentheses u to the power of negative 1 end exponent close parentheses space equals space minus space x squared space u to the power of negative 1 to the power of space end exponent plus space integral u to the power of negative 1 end exponent 2 x d x end style
begin mathsize 14px style I space equals space minus fraction numerator x squared over denominator x space tan x space plus 1 end fraction space plus space 2 integral fraction numerator x space d x over denominator x space tan x space plus 1 end fraction end style
begin mathsize 14px style I space equals space minus space fraction numerator x squared over denominator x space tan x space plus space 1 end fraction space plus space 2 space integral fraction numerator x space cos x space d x over denominator x space sin x space plus space cos x end fraction end style
differentiation of ( x sinx + cosx ) is determined below
 
d ( x sinx + cos x) = ( x cosx + sinx - sinx ) dx = x cosx dx
 
Then integration becomes 
 
begin mathsize 14px style I space equals space minus space fraction numerator x squared over denominator x space tan x space plus space 1 end fraction space plus space 2 space integral fraction numerator d left parenthesis space x space sin x space plus space cos x space right parenthesis over denominator x space sin x space plus space cos x end fraction end style
 
begin mathsize 14px style I space equals space minus space fraction numerator x squared over denominator x space tan x space plus space 1 end fraction space plus space 2 space log open parentheses x space sin x space plus space cos x close parentheses space plus space C end style
 
Hence we get, f(x) = 2 log ( x sinx + cosx )
Answered by Thiyagarajan K | 14 Oct, 2020, 09:50: PM

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