11.) by parts may be used........
1.) ∫xtanxsecxdx/(tanx-x)2
2.) ∫2dx/((x-5)+(x-7))√((x-5)(x-7)) = f(g(x)) +c, then
a) f(x)=sin^-1x , g(x) = √((x-5)(x-7))
b) f(x)=sin^-1x , g(x) = ((x-5)(x-7))
c) f(x)=tan^-1x , g(x) = √((x-5)(x-7))
d) f(x)=tan^-1x , g(x) =((x-5)(x-7))
3.) ∫dx/(x^7 - x^11) = (-1/x^6) - (1/2x2) - (1/4) g(x)
-
UP 0 DOWN 0 0 8
8 Answers
12.) ∫2dx/((x-5)+(x-7))√((x-5)(x-7))
= ∫∫dx/((x-6))√((x2-12x+35))
then x-6 = 1/t and so on.....
DONT DO IN THIS WAY....
THIS WAY INTEGRATION IS POSSIBLE....BUT OPTIONS NOT MATCHING :(
1357∫xtanxsecxdx/(tanx-x)2=∫(xsecx/tanx).(tan2x/(tanx-x)2dx
take (xsecx/tanx)=x/sinx as first function
and (tan2x/(tanx-x)2 as second function
1357for second put x-6=t
the integral becomes ∫2dt/2t√(t2-1=sec-1t+c=tan-1(t2-1)+c
=tan-1((x-5)(x-7))+c
so (d) is correct answer
1357the third one is ∫dx/x7(1-x4)
put x2=t
the integral becomes ∫dt/2t4(1-x2)
=(1/2)∫(t4-t4+t2-t2+1)/t4(1-x2)=(1/2)∫{1/(1-t2) + 1/t2 + 1/t4}dt
now solve it
1sorry bhaiya........
ans given for seconquestion is (c)........ may be wrong....
1and thank you for the last one.....i really didn have the hint........
thank you :)
1357ohh i did a mistake there
sec-1t+c=tan-1√(t2-1)+c
so the answer is (c) only