# 9 - Integrations Questions Answers

Joshi sir comment
Joshi sir comment

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

Solution by Joshi sir

How to inegrate ( In t) dt /(t-1)

Joshi sir comment

let lnt = x so dt/t = dx so dt = tdx = exdx

now ∫(ln t)dt/(t-1) = ∫xexdx/(ex-1) = ∫x(ex-1+1)dx/(ex-1)

now solve

Maths >> Calculus >> Integrations IIT JEE

∫√tanx

Joshi sir comment

let  √tanx = t

so sec2xdx/2√tanx = dt

so (1+t4)dx/2t = dt

so dx = 2tdt/(1+t4)

now solve

Maths >> Calculus >> Integrations IIT JEE

0π [cot-1x]dx

Joshi sir comment

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)

Maths >> Calculus >> Integrations Others
• what is the integration of cosx 2
Joshi sir comment

cosx2 = (cos2)x

so ∫ (cos2)x dx = (cos2) x / logecos2

Solution by Joshi sir

cosx2 = (cos2)x

so ∫ (cos2)x dx = (cos2) x / logecos2 + C            formula ∫axdx = ax/logea

Maths >> Calculus >> Integrations IIT JEE

Evaluate: 0ò11/{ (5+2x-2x2)(1+e(2-4x)) } dx

Joshi sir comment

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

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

we get 2I = 0ò11/(5+2x-2x2) dx

now solve this

Maths >> Calculus >> Integrations IIT JEE

Evaluate 0òx [x] dx .

Joshi sir comment

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

so 0òx [x] dx = 0ò1 0 dx + 1ò2 1 dx + 2ò3 2 dx + ..................... + [x]-1[x] [x]-1 dx + [x][x] dx

now solve it