7 - Integrations Questions Answers
1 . Iₙ = ∫ ( 1/ ( x² + a²)ⁿ )dx
How to do this by substituting x = a tan Α
2. Is this true
∫ f(x) d(kg(x)) = k ∫ f(x) d(g(x))
where k is a constant
How to inegrate ( In t) dt /(t-1)
let lnt = x so dt/t = dx so dt = tdx = e^{x}dx
now ∫(ln t)dt/(t-1) = ∫xe^{x}dx/(e^{x}-1) = ∫x(e^{x}-1+1)dx/(e^{x}-1)
now solve
∫√tanx
let √tanx = t
so sec^{2}xdx/2√tanx = dt
so (1+t^{4})dx/2t = dt
so dx = 2tdt/(1+t^{4})
now solve
∫_{0}π [cot^{-1}x]dx
i think the question will be ∫_{0}π [cotx]dx because limits are in angular terms.
by graph given below we get answer as -π/2
here in the graph same coloured shaded region are dx*(-1) always, because for every similar pair [+k]+[-k] = -1 , here k is a real number, so complete area = -1(π/2)
- what is the integration of cos^{x }2
cos^{x}2 = (cos2)^{x}
so ∫ (cos2)^{x} dx = (cos2) ^{x} / log_{e}cos2
cos^{x}2 = (cos2)^{x}
so ∫ (cos2)^{x} dx = (cos2) ^{x} / log_{e}cos2 + C formula ∫a^{x}dx = a^{x}/log_{e}a
Evaluate: _{0}ò^{1}1/{ (5+2x-2x^{2})(1+e^{(2-4x)}) } dx
let I = _{0}ò^{1}1/{ (5+2x-2x^{2})(1+e^{(2-4x)}) } dx
on using the property _{0}∫^{a} f(x) dx = _{0}∫^{a}f(a-x)dx
we get 2I = _{0}ò^{1}1/(5+2x-2x^{2}) dx
now solve this
Evaluate _{0}ò^{x} [x] dx .
integer nearst to x and less than x will be [x]
so _{0}ò^{x} [x] dx = _{0}ò^{1} 0 dx + _{1}ò^{2} 1 dx + _{2}ò^{3} 2 dx + ..................... + _{[x]-1}∫^{[x] }[x]-1 dx + _{[x]}∫^{x }[x] dx
now solve it