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AllQuestion and Answers: Page 1682 |
∫_(−π/2) ^(+π/2) (√(cos^(2n−1) x−cos^(2n+1) x))dx =[−((2cos^((2n+1)/2) x)/(2n+1))]_(−π/2) ^(+π/2) =0? What is the mistake in above? ∫_(−π/2) ^(+π/2) (√(cos^(2n−1) x−cos^(2n+1) x))dx =2∫_0 ^(π/2) (√(cos^(2n−1) x−cos^(2n+1) x))dx =(4/(2n+1)) (this is correct answer) |
show that C_(n−1) ^(n+r−1) =C_r ^(n+r−1) |
4 couples are to take a photograph with a newly wedded couple in a wedding party.In how many ways can this be done if: i)the celebrated couple must stand in the middle ii)each couple must stand next to each other iii)the celebrated couple must not stand next to each other |
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find f(x)=∫_0 ^∞ ((arctan(x(t +(1/t))))/(1+t^2 ))dt |
let f(x)= e^(−(√(1+2x))) developp f at integr serie . |
let f(x)=x arctan(1+e^(−x) ) developp f at intrgr serie . |
let f(x)= ((artan(x+1))/(1+2x)) developp f at integr serie . |
find the value of f(x) = ∫_0 ^π ((cosx)/(1+2sin(2x)))dx |
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let f(x) = (e^(−x) /(2+x)) developp f at integr serie. |
lim_ _(x→∞) ((ln x)/x) = ? You can only use series expansion / sandwich theorem! |
Find ∫ Sin^6 x dx |
Find range of y=(x/((x−1)(x−2))) . |
solve for x 4x=2^x |
f(x)=cos(x) g(x)=2^x f:R→R g:R→R^+ [f(x)]^2 +[f(π/2−x)]^2 =((log_2 g(x))/x);x≠0 f(g(x)x)=[f(g(x−1)x)]^2 +[f(π/2−g(x−1)x)]^2 find f and g |
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find lim_(n→+∞) ((n^p sin^2 (n!))/n^(p+1) ) with0<p<1 . |
let A(x)= ∫_0 ^1 ln(1+ix^2 )dx find a simple form of f(x) (x∈R) |
let f(x)= ln(1+ix^2 ) 1) extrsct Re(f(x)) and Im(f(x)) 2) developp f at integr serie 3) calculate f^′ (x) by two methods |
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lim_(x→0) ((1/x) − ((ln^(1000) (1 + x))/x^(1001) )) |
Pg 1677 Pg 1678 Pg 1679 Pg 1680 Pg 1681 Pg 1682 Pg 1683 Pg 1684 Pg 1685 Pg 1686 |