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

 

Asked By: LUFFY
is this question helpfull: 2 0 submit your answer
Solution by Joshi sir

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

Asked By: PARTHASARATHY
is this question helpfull: 1 0 submit your answer
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

Asked By: BONEY HAVELIWALA
is this question helpfull: 4 0 submit your answer
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

Asked By: NIKHIL VARSHNEY
is this question helpfull: 2 1 submit your answer
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
Asked By: VIKRANT KUMAR
is this question helpfull: 3 0 submit your answer
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

Asked By: KAMAL
is this question helpfull: 7 0 submit your answer
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 .

Asked By: KAMAL
is this question helpfull: 6 0 submit your answer
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

 

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