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

Explanation:

$\mathrm{I}=\int \frac{\sqrt{\tan \mathrm{x}}}{\sin \mathrm{x}\cos \mathrm{x}}\mathrm{d}\mathrm{x}$

   = $\int \frac{\sqrt{\tan x}}{\tan x{\cos }^{2}x}dx$

  = $\int \frac{\sqrt{\tan x}}{\tan x}{\sec }^{2}xdx$

  = $\int \frac{1}{\sqrt{\tan x}}{\sec }^{2}xdx$

Let tan x = t² ⇒ sec²x dx = 2tdt

So, I = $\int \frac{2t}{t}dt$

        = 2t + c

      = $2\sqrt{\tan \mathrm{x}}+\mathrm{c}$

Option (3) is true.

Concept Used:

d(u × v) = u × dv + v × du

Calculation:

$\int \frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{x}+1)}{{\cos }^2(\mathrm{x}{\mathrm{e}}^{\mathrm{x}})}\mathrm{d}\mathrm{x}$

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

∴​ $\int \frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{x}+1)}{{\cos }^2(\mathrm{x}{\mathrm{e}}^{\mathrm{x}})}\mathrm{d}\mathrm{x}$ = tan(xe^x) + c

Formula Used:

cos2x = cos²x - sin²x

tanx = sinx/cosx

$\int \frac{1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}$ = sin^-1x + c

Calculation:

Let,

I = $\int \frac{\sec \mathrm{x}}{\surd \cos 2\mathrm{x}}\mathrm{d}\mathrm{x}$

⇒ I =  $\int \frac{\sec \mathrm{x}}{\sqrt{{\cos }^2\mathrm{x}-\mathrm{s}\mathrm{i}{\mathrm{n}}^2\mathrm{x}}}\mathrm{d}\mathrm{x}$

⇒ I = $\int \frac{\sec \mathrm{x}}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}\sqrt{1-\mathrm{t}\mathrm{a}{\mathrm{n}}^2\mathrm{x}}}\mathrm{d}\mathrm{x}$

⇒ I = $\int \frac{{\sec }^2\mathrm{x}}{\sqrt{1-\mathrm{t}\mathrm{a}{\mathrm{n}}^2\mathrm{x}}}\mathrm{d}\mathrm{x}$

Let tanx = t 

Differential with respect to t

⇒ I = (sec²x) dx = dt

⇒ I = $\int \frac{1}{\sqrt{1-{\mathrm{t}}^2}}\mathrm{d}\mathrm{t}$

⇒ sin^-1t + c

∴ sin^-1(tanx) + c

Solution

⇒ $\int \frac{dx}{1+\sin x}$

Dividing and multiplying the term by its conjugate.

⇒ $\int \frac{dx}{1+\sin x}$ × $\frac{1-sinx}{1+sinx}$dx

⇒ $\int \frac{1-sinx}{1-{\sin }^{2}x}$  ...(1 - sin²x = cos²x)

⇒ $\int \frac{1-sinx}{co{s}^{2}x}$ = $\int \frac{1}{co{s}^{2}x}-\frac{sinx}{co{s}^{2}x}$ 

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:

$\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x+C$

2 sin x cos x =sin 2x

Calculation:

Let $I={\displaystyle \int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$ . . .(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)

$I={\displaystyle \int \frac{dt}{\sqrt{1-t^2}}}$

$I={\sin }^{-1}t+C$

$I={\sin }^{-1}(\sin x-\cos x)+C$

Concept:

$\int \frac{1}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}={\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$

Calculation:

I = $\int \frac{1}{\sqrt{16-25{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}$

= $\int \frac{1}{\sqrt{16-(5\mathrm{x}{)}^2}}\mathrm{d}\mathrm{x}$

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = $\frac{\mathrm{d}\mathrm{t}}{5}$

Now,

I = $\frac{1}{5}\int \frac{1}{\sqrt{4^2-{\mathrm{t}}^2}}\mathrm{d}\mathrm{t}$

= $\frac{1}{5}{\sin }^{-1}\left(\frac{\mathrm{t}}{4}\right)$ + c

= $\frac{1}{5}{\sin }^{-1}\left(\frac{5\mathrm{x}}{4}\right)$ + c

Concept:

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation:

I = $\int \sqrt{2\mathrm{x}+3}\mathrm{d}\mathrm{x}$

Let 2x + 3 = t²

Differenating with respect to x, we get

⇒ 2dx = 2tdt

⇒ dx = tdt

Now,

I = $\int \sqrt{{\mathrm{t}}^2}\times \mathrm{t}\mathrm{d}\mathrm{t}$

= $\int {\mathrm{t}}^2\mathrm{d}\mathrm{t}$

= $\frac{{\mathrm{t}}^3}{3}+\mathrm{c}$

∵ 2x + 3 = t2

⇒  (2x + 3)^1/2 = t

⇒ (2x + 3)³/2 = t³

⇒ I = $\frac{(2\mathrm{x}+3{)}^{3/2}}{3}+\mathrm{c}$

Concept:

$\int \sin \mathrm{x}\mathrm{d}\mathrm{x}=-\cos \mathrm{x}+\mathrm{c}$

Calculation:

I = $\int \sin 5\mathrm{x}\mathrm{d}\mathrm{x}$

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = $\frac{\mathrm{d}\mathrm{t}}{5}$

Now,

I = $\frac{1}{5}\int \sin \mathrm{t}\mathrm{d}\mathrm{t}$

= $\frac{1}{5}(-\cos \mathrm{t})+\mathrm{c}$

= $\frac{-\cos 5\mathrm{x}}{5}+\mathrm{c}$

Concept: 

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}$

$\int \mathrm{f}(\mathrm{x})\mathrm{d}{\mathrm{x}}^2$ = $\int (1+{\mathrm{x}}^2+{\mathrm{x}}^4)\mathrm{d}({\mathrm{x}}^2)$      ....(i)

Calculation:

Let, x² = u

From equation (i)

​$\int \mathrm{f}(\mathrm{x})\mathrm{d}{\mathrm{x}}^2$ = $\int (1+\mathrm{u}+{\mathrm{u}}^2)\mathrm{d}\mathrm{u}$

⇒ u + $\frac{{\mathrm{u}}^2}{2}$ + $\frac{{\mathrm{u}}^3}{3}$+ C

Now putting the value of u,

​⇒ $\int \mathrm{f}(\mathrm{x})\mathrm{d}{\mathrm{x}}^2$ = x² +​ $\frac{{\mathrm{x}}^4}{2}$ + $\frac{{\mathrm{x}}^6}{3}$ + C

∴ The required integral is x2 +​ $\frac{{\mathrm{x}}^4}{2}$ + $\frac{{\mathrm{x}}^6}{3}$ + C.

Concept:

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation:

I = $\int \sqrt{4\mathrm{x}-3}\mathrm{d}\mathrm{x}$

Let 4x - 3 = t²

Differenating with respect to x, we get

⇒ 4dx = 2tdt

⇒ dx = $\frac{\mathrm{t}}{2}$dt

Now,

I = $\int \sqrt{{\mathrm{t}}^2}\times \frac{\mathrm{t}}{2}\mathrm{d}\mathrm{t}$

= $\frac{1}{2}\int {\mathrm{t}}^2\mathrm{d}\mathrm{t}$

= $\frac{{\mathrm{t}}^3}{6}+\mathrm{c}$

= $\frac{(4\mathrm{x}-3{)}^{3/2}}{6}+\mathrm{c}$

Concept:

Integral property:

• ∫ xn dx = $\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$+ C ; n ≠ -1
• $\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\ln \mathrm{x}$ + C

Calculation:

Let I = $\int \frac{\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}$

I = $\frac{1}{2}\int \frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}$

Let 1 + x2 = t

⇒ 2x dx = dt

I = $\frac{1}{2}\int \frac{1}{\mathrm{t}}\mathrm{d}\mathrm{t}$

= $\frac{1}{2}\log \mathrm{t}+\mathrm{c}$

= $\frac{1}{2}\log (1+{\mathrm{x}}^2)+\mathrm{c}$

= $\log \sqrt{(1+{\mathrm{x}}^2)}+\mathrm{c}$              [∵ n log m = log mⁿ]

Calculation:

Given:$I_1={\displaystyle {\int }_e^{e^2}{\displaystyle \frac{dx}{\log x}}}$ and $I_2={\displaystyle {\int }_1^2{\displaystyle \frac{e^x}{x}}dx}$

⇒ $I_1={\displaystyle {\int }_e^{e^2}{\displaystyle \frac{dx}{\log x}}}$ 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 = ${\displaystyle {\int }_1^2}$(e^z dz) / z =${\displaystyle {\int }_1^2}$(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 = $\int \frac{2}{\sin 2\mathrm{x}.\log \left(\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{x}\right)}$         ....(1)

Take log (tan x) = t

$\frac{1}{\tan \mathrm{x}}(\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x})\mathrm{d}\mathrm{x}=\mathrm{d}\mathrm{t}$

⇒ $\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}.\mathrm{c}\mathrm{o}{\mathrm{s}}^2\mathrm{x}}\mathrm{d}\mathrm{x}=\mathrm{d}\mathrm{t}$

⇒ $\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}.\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}\mathrm{d}\mathrm{x}=\mathrm{d}\mathrm{t}$

⇒ dx = sin x.cos x dt

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

Now, I = $\int \frac{2}{2\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}.\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}.\mathrm{t}}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}.\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}\mathrm{d}\mathrm{t}$

= ∫ $\frac{1}{\mathrm{t}}$dt

= log t + c

= log [log(tan x) ]+ c

Concept:

$\int \frac{1}{1+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\mathrm{x}+\mathrm{c}$

${\mathrm{x}}^{-1}=\frac{1}{\mathrm{x}}$

Calculation:

Let, I = $\int \frac{1}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}\mathrm{d}\mathrm{x}$

= $\int \frac{1}{{\mathrm{e}}^{\mathrm{x}}+\frac{1}{{\mathrm{e}}^{\mathrm{x}}}}\mathrm{d}\mathrm{x}$                (∵ ${\mathrm{x}}^{-1}=\frac{1}{\mathrm{x}}$)

= $\int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{2\mathrm{x}}+1}\mathrm{d}\mathrm{x}$

Now, let e^x = t

⇒ ex dx = dt

∴ I = $\int \frac{\mathrm{d}\mathrm{t}}{{\mathrm{t}}^2+1}$

= $\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\mathrm{t}+\mathrm{c}$             (∵ $\int \frac{1}{1+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\mathrm{x}+\mathrm{c}$)

= $\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}({\mathrm{e}}^{\mathrm{x}})+\mathrm{c}$         (∵ ex = t)

Hence, option (4) is correct. 

Concept:

$\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\log |\mathrm{x}|+\mathrm{c}$

Calculation:

I = ∫ cot 2x dx

$=\int \frac{\cos 2\mathrm{x}}{\sin 2\mathrm{x}}\mathrm{d}\mathrm{x}$

Let sin 2x = t

Differentiating with respect to x, we get

⇒ 2 cos 2x dx = dt

⇒ cos 2x dx = $\frac{\mathrm{d}\mathrm{t}}{2}$

$\mathrm{I}=\frac{1}{2}\int \frac{1}{\mathrm{t}}\mathrm{d}\mathrm{t}$

= $\frac{1}{2}\log |\mathrm{t}|+\mathrm{c}$

= $\frac{1}{2}\log |\sin 2\mathrm{x}|+\mathrm{c}$