Integration using Substitution MCQ Quiz - Objective Question with Answer for Integration using Substitution - Download Free PDF

Explanation:

   = 

  = 

  = 

Let tan x = t² ⇒ sec²x dx = 2tdt

So, I = 

        = 2t + c

      = 

Option (3) is true.

Concept Used:

d(u × v) = u × dv + v × du

Calculation:

Let xe^x = t 

Differential with respect to t

(xe^x + e^x) dx = dt 

ex (x + 1) dx = dt 

Now, ∫(1/cos²t) dt

⇒ ∫sec²t dt 

⇒ tant + c

∴​  = tan(xe^x) + c

Formula Used:

cos2x = cos²x - sin²x

tanx = sinx/cosx

 = sin^-1x + c

Calculation:

Let,

I = 

⇒ I =  

⇒ I = 

⇒ I = 

Let tanx = t 

Differential with respect to t

⇒ I = (sec²x) dx = dt

⇒ I = 

⇒ sin^-1t + c

∴ sin^-1(tanx) + c

Solution

⇒ 

Dividing and multiplying the term by its conjugate.

⇒  × dx

⇒   ...(1 - sin²x = cos²x)

⇒  =  

Since, (1/cos²x = sec²x and sin/cos²x = tan x × sec x) 

⇒ ∫(sec²x - tan x sec x) dx

⇒ ∫sec²x dx - ∫tan x sec x dx

⇒ tan x - sec x + c

The correct option is 2.

Formula Used:

2 sin x cos x =sin 2x

Calculation:

Let  . . .(1)

Now, Put sin x - cos x = t

⇒​ ( sin x + cos x ) dx = dt

and (sin x - cos x)² = t²

⇒ 1 - 2 sin x cos x =  t2

⇒ 1 - sin 2x =  t2

⇒ 1 - t2  =  sin 2x

Substituting all the values in (1)

Concept:

Calculation:

I = 

= 

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = 

Now,

I = 

=  + c

=  + c

Concept:

Calculation:

I = 

Let 2x + 3 = t²

Differenating with respect to x, we get

⇒ 2dx = 2tdt

⇒ dx = tdt

Now,

I = 

= 

= 

∵ 2x + 3 = t2

⇒  (2x + 3)^1/2 = t

⇒ (2x + 3)³/2 = t³

⇒ I = 

Concept:

Calculation:

I = 

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = 

Now,

I = 

= 

= 

Concept: 

 =       ....(i)

Calculation:

Let, x² = u

From equation (i)

​ = 

⇒ u +  + + C

Now putting the value of u,

​⇒  = x² +​  +  + C

∴ The required integral is x2 +​  +  + C.

Concept:

Calculation:

I = 

Let 4x - 3 = t²

Differenating with respect to x, we get

⇒ 4dx = 2tdt

⇒ dx = dt

Now,

I = 

= 

= 

= 

Concept:

Integral property:

• ∫ xn dx = + C ; n ≠ -1
•  + C

Calculation:

Let I = 

I = 

Let 1 + x2 = t

⇒ 2x dx = dt

I = 

= 

= 

=               [∵ n log m = log mⁿ]

Calculation:

Given: and 

⇒  put log x = z

Such that x = e^z

Such that dx = e^z dz 

when x = e, z = loge

x = e², z = log e² = 2 log e = z 

Such that I₁ = (e^z dz) / z =(e^x/z) dx = I₂

Such that I₁ = I₂

⇒ I1 - I2 = 0 

Concept:

sin 2x = 2sin x cos x

∫(1/x)dx = log x + c

∫tanx dx = sec²x + c  

Calculation:

Let I =          ....(1)

Take log (tan x) = t

⇒ 

⇒ 

⇒ dx = sin x.cos x dt

Putting the value of log (tan x) and dx in equation (i)

Now, I = 

= ∫ dt

= log t + c

= log [log(tan x) ]+ c

Concept:

Calculation:

Let, I = 

=                 (∵ )

= 

Now, let e^x = t

⇒ ex dx = dt

∴ I = 

=              (∵ )

=          (∵ ex = t)

Hence, option (4) is correct. 

Concept:

Calculation:

I = ∫ cot 2x dx

Let sin 2x = t

Differentiating with respect to x, we get

⇒ 2 cos 2x dx = dt

⇒ cos 2x dx = 

= 

=