INDEFINITE INTEGRALS USING PROPERTIES

Question 1 :

Integrate the following with respect to x:

e^x (tan x + log sec x)

Solution :

  =  ∫e^x (tan x + log sec x) dx

Let f(x)  =  log sec x

Differentiating with respect to "x"

  f'(x)  =  (1/sec x) (sec x tan x) dx

  f'(x)  =  tan x

 ∫e^x (tan x + log sec x) dx  =   ∫e^x (f(x) + f'(x)) dx

   =  e^x f(x) + c 

   =  e^x log sec x + c 

Question 2 :

Integrate the following with respect to x:

e^x [(x - 1)/2x²]

Solution :

  =  ∫e^x [(x - 1)/2x²] dx

=  ∫e^x [(x/2x²) - (1/2x²)] dx

=  ∫e^x [(1/2x) - (1/2x²)] dx

Let f(x)  =  1/2x then f'(x)  =  -1/2x²

Hence ∫e^x [(1/2x) - (1/2x²)] dx  is in the form ∫e^x (f(x) + f'(x)) dx

  =  e^x f(x)

  =  e^x (1/2x)

  =  e^x/2x

Question 3 :

Integrate the following with respect to x:

e^x sec x (1 + tan x)

Solution :

  =  ∫e^x sec x (1 + tan x) dx

  =  ∫e^x (sec x  + sec x tan x) dx

Let f(x)  =  sec x, then f'(x)  =  sec x tan x

Hence ∫e^x (sec x  + sec x tan x) dx is in the form ∫e^x (f(x) + f'(x)) dx

  =  e^x f(x)

  =  e^x sec x + c

Question 4 :

Integrate the following with respect to x:

e^x (2 + sin 2x) / (1 + cos 2x)

Solution :

  =  ∫e^x [(2 + sin 2x) / (1 + cos 2x)] dx

sin 2x  =  2 sin x cos x

1 + cos 2x  =  2 cos²x

  =  ∫e^x [(2 + 2 sin x cos x) / 2 cos²x] dx

  =  ∫e^x [(2 / 2 cos²x)  + (2 sin x cos x / 2 cos²x)] dx

  =  ∫e^x [(1/cos²x)  + (sin x / cosx)] dx

  =  ∫e^x [sec²x  + tan x] dx  ------(1)

Let f(x)  =  tan x then f'(x)  =  sec²x

Hence (1) is in the form 

  =  ∫e^x [f(x) + f'(x)] dx

=  e^x f(x) + c

=  e^x tan x + c

Question 5 :

Integrate the following with respect to x:

e^tan^-1x [(1 + x + x²) / (1 + x²)]

Solution :

In order to compare the given question with the form ∫e^x [f(x) + f'(x)] dx, we must have power "x" to the power of "e".

Let t = tan^-1x

x  =  tan t

dt  =  1/1 + x²

∫e^tan^-1x [(1 + x + x²) / (1 + x²)] dx

 =  ∫e^t (1 + tan t + tan²t) dt

Here 1 + tan²t  =  sec²t

 =  ∫e^t (tan t + sec²t) dt

f(x)  =  tan t then f'(x)  =  sec²t

=  e^t tan t + c

=  e^tan^-1x tan (tan^-1x) + c

=  x e^tan^-1x + c

Question 6 :

Integrate the following with respect to x:

log x / (1 + log x)²

Solution :

=  ∫ [log x / (1 + log x)²] dx

Let t = log x ==> e^t  =  x

Differentiating with respect to "x", we get

e^t dt =  dx

=  ∫ [t/(1 + t)²] e^t dt

=  ∫ e^t [(t + 1 - 1) /(1 + t)²] dt

=  ∫ e^t [((1 + t)/(1 + t)²) - 1/(1 + t)²] dt

=  ∫ e^t [1/(1 + t) - 1/(1 + t)²] dt

If f(t)  =  1/(1 + t) then f'(t)  =  -1/(1 + t)²

  =  e^t f(t) + c

  = e^t  (1/(1 + t)) + c

  = x/(1 + log x) + c

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