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

Explanation:

\(\rm I=\int\frac{\sqrt{\tan x}}{\sin x \cos x}dx\)

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

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

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

Let tan x = t² ⇒ sec²x dx = 2tdt

So, I = \(\int\frac{2t}{t}dt\)

        = 2t + c

      = \(\rm 2\sqrt{\tan x}+c\)

Option (3) is true.

Concept Used:

d(u × v) = u × dv + v × du

Calculation:

\(\rm ∫ \frac{e^x(x+1)}{\cos^2(xe^x)}dx\)

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

∴​ \(\rm ∫ \frac{e^x(x+1)}{\cos^2(xe^x)}dx\) = tan(xe^x) + c

Formula Used:

cos2x = cos²x - sin²x

tanx = sinx/cosx

\(\rm \int\frac{1}{\sqrt{1-x^2}}dx \) = sin^-1x + c

Calculation:

Let,

I = \(\rm ∫\frac{\sec x}{√{\cos 2x}}dx\)

⇒ I =  \(\rm \int\frac{\sec x}{\sqrt{\cos^2x-sin^2x}}dx\)

⇒ I = \(\rm \int\frac{\sec x}{cosx\sqrt{1-tan^2x}}dx\)

⇒ I = \(\rm \int\frac{\sec^2 x}{\sqrt{1-tan^2x}}dx\)

Let tanx = t 

Differential with respect to t

⇒ I = (sec²x) dx = dt

⇒ I = \(\rm \int\frac{1}{\sqrt{1-t^2}}dt\)

⇒ sin^-1t + c

∴ sin^-1(tanx) + c

Solution

⇒ \(\int {\frac{{dx}}{{1 + \sin x}}} \)

Dividing and multiplying the term by its conjugate.

⇒ \(\int {\frac{{dx}}{{1 + \sin x}}} \) × \(1 - sinx \over1 + sinx\)dx

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

⇒ \(\int {\frac{{1-sinx}}{{cos^2 x}}} \) = \(\int {\frac{{1}}{{cos^2 x}}}- {\frac{{sinx}}{{cos^2 x}}}\) 

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:

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

Calculation:

I = \(\rm \int \frac {1}{\sqrt{16-25x^2}}dx\)

= \(\rm \int \frac {1}{\sqrt{16-(5x)^2}}dx\)

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = \(\rm \frac {dt}{5}\)

Now,

I = \(\rm \frac {1}{5}\int \frac {1}{\sqrt{4^2-t^2}} dt\)

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

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

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1} +c\)

Calculation:

I = \(\rm \int \sqrt{2x+3}\;dx\)

Let 2x + 3 = t²

Differenating with respect to x, we get

⇒ 2dx = 2tdt

⇒ dx = tdt

Now,

I = \(\rm \int \sqrt{t^2}\; \times tdt\)

= \(\rm \int t^2 \;dt\)

= \(\rm \frac {t^3}{3} + c\)

∵ 2x + 3 = t2

⇒  (2x + 3)^1/2 = t

⇒ (2x + 3)³/2 = t³

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

Concept:

\(\rm \int \sin x \; dx = -\cos x + c\)

Calculation:

I = \(\rm \int \sin 5x\;dx \)

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = \(\rm \frac{dt}{5} \)

Now,

I = \(\rm \frac 1 5 \int \sin t\;dt \)

= \(\rm \frac 1 5 (-\cos t) + c\)

= \(\rm \frac{-\cos 5x}{5} + c\)

Concept: 

\(\rm \int x^{n}\space dx = \frac{x^{n + 1}}{n + 1} + C\)

\(\rm \int f(x) \space dx^2\) = \(\rm \int (1 + x^{2} + x^{4}) \space d(x^2)\)      ....(i)

Calculation:

Let, x² = u

From equation (i)

​\(\rm \int f(x) \space dx^2\) = \(\rm \int (1 + u + u^{2}) \space du\)

⇒ u + \(\rm \frac{u^{2}}{2}\) + \(\rm \frac{u^{3}}{3}\)+ C

Now putting the value of u,

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

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

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1} +c\)

Calculation:

I = \(\rm \int \sqrt{4x-3}\;dx\)

Let 4x - 3 = t²

Differenating with respect to x, we get

⇒ 4dx = 2tdt

⇒ dx = \(\rm \frac t 2\)dt

Now,

I = \(\rm \int \sqrt{t^2}\; \times \frac t 2 dt\)

= \(\frac12 \rm \int t^2 \;dt\)

= \(\rm \frac {t^3}{6} + c\)

= \(\rm \frac {(4x-3)^{3/2}}{6} + c\)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C

Calculation:

Let I = \(\rm \int{x\over1+x^2} \;dx\)

I = \(\rm \frac 12 \int{2x\over1+x^2} \;dx\)

Let 1 + x2 = t

⇒ 2x dx = dt

I = \(\rm \frac 12 \int{1\over t } dt\)

= \(\rm \frac 12 \log t + c\)

= \(\rm \frac 12 \log (1 + x^2) + c\)

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

Calculation:

Given:\(I_1 = \displaystyle\int_e^{e^2} \dfrac{dx}{\log x}\) and \(I_2 = \displaystyle\int_1^2 \dfrac{e^x}{x} dx\)

⇒ \(I_1 = \displaystyle\int_e^{e^2} \dfrac{dx}{\log x}\) 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 = \(\smallint \frac{2}{{\sin 2{\rm{x}}.\log \left( {{\rm{tanx}}} \right)}}\)         ....(1)

Take log (tan x) = t

\(\rm \frac1 {\tan x}(se{c^2}x)dx = dt\)

⇒ \( \frac{{{\rm{cosx}}}}{{{\rm{sinx}}.{\rm{\;co}}{{\rm{s}}^2}{\rm{x}}}}{\rm{\;dx}} = {\rm{dt}}\)

⇒ \( \frac{1}{{{\rm{sinx}}.{\rm{cosx}}}}{\rm{dx}} = {\rm{dt}}\)

⇒ dx = sin x.cos x dt

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

Now, I = \(\rm \smallint \frac{2}{{2sinx.cosx.t}}\;sinx.cosx\;dt\)

= ∫ \(\rm \frac 1 t\)dt

= log t + c

= log [log(tan x) ]+ c

Concept:

\(\rm \int \frac{1}{1+x^2}dx=tan^{-1}x+c\)

\(\rm x^{-1}=\frac1x\)

Calculation:

Let, I = \(\rm \int \frac{1}{e^x+e^{-x}}dx\)

= \(\rm \int \frac{1}{e^x+\frac{1}{e^{x}}}dx\)                (∵ \(\rm x^{-1}=\frac1x\))

= \(\rm \int \frac{e^x}{e^{2x}+{1}}dx\)

Now, let e^x = t

⇒ ex dx = dt

∴ I = \(\rm \int \frac{dt}{t^2+{1}}\)

= \(\rm tan^{-1}t+c\)             (∵ \(\rm \int \frac{1}{1+x^2}dx=tan^{-1}x+c\))

= \(\rm tan^{-1}(e^x)+c\)         (∵ ex = t)

Hence, option (4) is correct. 

Concept:

\(\rm \int \frac{1}{x}dx = \log |x| + c\)

Calculation:

I = ∫ cot 2x dx

\(=\rm \int \frac{\cos 2x}{\sin 2x}dx\)

Let sin 2x = t

Differentiating with respect to x, we get

⇒ 2 cos 2x dx = dt

⇒ cos 2x dx = \(\rm \frac {dt}{2}\)

\(\rm I =\rm \frac 1 2\int \frac{1}{t}dt\)

= \(\rm \frac 1 2 \log |t| + c\)

= \(\rm \frac 1 2 \log |\sin 2x| + c\)