18 - Integrations Questions Answers

let lnt = x so dt/t = dx so dt = tdx = e^xdx

now ∫(ln t)dt/(t-1) = ∫xe^xdx/(e^x-1) = ∫x(e^x-1+1)dx/(e^x-1)

now solve 

let  √tanx = t

so sec²xdx/2√tanx = dt

so (1+t⁴)dx/2t = dt

so dx = 2tdt/(1+t⁴)

now solve

i think the question will be ∫₀π [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)

cos^x2 = (cos2)^x

so ∫ (cos2)^x dx = (cos2) ^x / log_ecos2

cos^x2 = (cos2)^x

so ∫ (cos2)^x dx = (cos2) ^x / log_ecos2 + C            formula ∫a^xdx = a^x/log_ea

let I = ₀ò¹1/{ (5+2x-2x²)(1+e^(2-4x)) } dx

on using the property ₀∫^a f(x) dx = ₀∫^af(a-x)dx

we get 2I = ₀ò¹1/(5+2x-2x²) dx

now solve this 

integer nearst to x and less than x will be [x]

so ₀ò^x [x] dx = ₀ò¹ 0 dx + ₁ò² 1 dx + ₂ò³ 2 dx + ..................... + _[x]-1∫^[x] [x]-1 dx + _[x]∫^x [x] dx

now solve it