Evaluate using Substitution MCQ Quiz in বাংলা - Objective Question with Answer for Evaluate using Substitution - বিনামূল্যে ডাউনলোড করুন [PDF]

Formula used:

• \(\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx\)
• 1 + tan²θ = sec²θ 
• \(tan(A-B)=\frac{tan\ A-tan\ B}{1+tan\ A\times tan\ B}\)
• log(m/n) = log m - log n
• \(\int x^ndx = \frac{x^{n+1}}{n+1}+C\)

Calculation:

Let,

\(I=\displaystyle \int_0^1 \frac{\log (1+x)}{1+x^2}dx\)

Put x = tan θ

⇒ dx = sec²θ dθ 

When x = 0 then θ = 0

When x = 1 then θ = π/4 

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} \frac{\log (1+tan\ θ)}{1+\tan^2θ}sec^2θ \ dθ\)

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} \frac{\log (1+tan\ θ)}{sec^2θ}sec^2θ \ dθ\)

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log (1+tan\ θ)} \ dθ\)    ----(1)

Using the property

\(\int_{0}^{a}f(x)dx=n\int_{0}^{a}f(a-x)dx\)

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log (1+tan\ (π/4 - θ))} \ dθ\)

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log (1+\frac{\ tan\frac{π}{4} - tanθ}{1+tan\frac{π}{4} \times tanθ}}) \ dθ\)

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log (1+\frac{1 - tanθ}{1+1 \times tanθ}}) \ dθ\)     [∵ tan(π/4) = 1]

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log (\frac{1 + tan \ \theta+ 1 - tanθ}{1+ tanθ}}) \ dθ\)

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log (\frac{2}{1+ tanθ}} )\ dθ\)    

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log {2}-log({1+ tanθ}} )\ dθ\)

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log {2}-\displaystyle \int_0^{\frac{π}{4}}log({1+ tanθ}} )\ dθ\)

Using the equation (1)

\(\Rightarrow I=\displaystyle \int_0^{\frac{π}{4}} {\log {2}}\ dθ-I\)

\(\Rightarrow 2I=\log {2}\displaystyle \int_0^{\frac{π}{4}} {}dθ\)

\(\Rightarrow 2I=\log {2}\times (θ)_0^{\frac{π}{4}} \)

\(\Rightarrow 2I=\log {2}\times {\frac{π}{4}} \)

\(\therefore I={\frac{π}{8}} \log {2} \)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to be determined.
• Substitution is done and the integral function is then integrated.
• Finally, the initial variable t, to be returned.

Calculation:

I = \(\rm \int{\ln x\over x}dx\)

Let ln x = t ⇒\(\rm1\over x\)dx = dt

I = \(\rm \int{t}dt\)

I = \(\rm {t^2\over 2}\)

I = \(\rm {(\ln x)^2}\)

Putting the limits 

I = \(\frac{1}{2} \times \rm \left[(\ln x)^2\right]_1^4\)

I = \(\frac{1}{2} \times \rm \left[(\ln 4)^2- (\ln 1)^2\right]\)

I = \(\frac{1}{2} \times \rm (\ln 4)^2 = {({ln \ 4})^2\over 2} \)

Calculation:

Let I = \(\rm \int_0^{\frac{π}{4}}\sin^3 θ\ dθ\)

⇒ I = \(\rm \int_0^{\frac{π}{4}}\sin^2 θ.\sin θ\ dθ\)

⇒ I = \(\rm \int_0^{\frac{π}{4}}(1 -\cos^2 θ).\sin θ\ dθ\)

Let cos θ = u → -sin θ dθ = du

and when θ = 0, u = 1 and when θ = π/4, u = 1/√2

So by substituting these values in I,

⇒ I = \(\rm \int_1^{\frac{1}{\sqrt 2}}(1 -u^2).(-du)\)

⇒ I = \(\rm \int_1^{\frac{1}{\sqrt 2}}(u^2-1)du\)

⇒ I = \(\rm ({u^3 \over 3} - u)|_1^{\frac{1}{\sqrt 2}}\)

⇒ I = \(\rm ({1\over 3}\times{1 \over 2\sqrt 2} - {1 \over \sqrt 2} )- ({1\over 3} - 1)\)

⇒ I = \(\rm \frac{2}{3}-\frac{5}{6\sqrt2}\)

∴ The correct option is (1).

Concept:

Formulae

• tan-1x + cot-1x =  \(\frac{\pi}{2}\) 
• \(tan^{-1}\frac{1}{x}=cot^{-1}x\)
• tan-1x - tan-1y = \(tan^{-1}(\frac{x-y}{1+x\times y})\)
• \(\displaystyle \int\frac {1}{(x^2 + 1)} dx=tan^{-1}x + c\)

Calculation:

Let I = \(\displaystyle \int_0^1\frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}dx \)

⇒ I = \(\displaystyle \int_0^1\frac{2(x^2+2x+2)-(x + 1)}{(x+1)(x^2+2x+2)}dx \)

⇒ I = \(\displaystyle \int_0^1\frac {2}{(x + 1)} - \frac{1}{(x^2+2x+1+1)}dx \)

⇒ I = \(\displaystyle 2 \int_0^1\frac {1}{(x + 1)} dx- \int_0^1 \frac{1}{(x+1)^2+1}dx \)

⇒ I = \(2\left[ln ∣ x+1∣\right]_0^1-\left[tan^{-1}(x+1)\right]_0^1 \)

⇒ I = 2(ln 2 - ln 1) - (tan^-12 - tan^-11)      ------(i)

⇒ I = ln 2² - tan-12 + \(\frac{\pi}{4}\)

So, Statement I is correct.

⇒ I = ln 4 - tan-12 + \(\frac{\pi}{4}\)

As tan-12 + cot-12 =  \(\frac{\pi}{2}\) ⇒ tan-12 = \(\frac{\pi}{2}\) - cot-12

⇒ I = ln 4 - (\(\frac{\pi}{2}\) - cot-12) + \(\frac{\pi}{4}\)

⇒ I = ln 4 + cot-12 - \(\frac{\pi}{4}\)

So, Statement II is incorrect.

Again from Equation (i), we have 

I = 2(ln 2 - ln 1) - (tan-12 - tan-11)

⇒ I = 2 ln 2 - \(tan^{-1}(\frac{2-1}{1+2\times1})\)

⇒ I = 2 ln 2 - \(tan^{-1}\frac{1}{3}\)

⇒ I = 2 ln 2 - cot^-13

So, Statement III is correct

∴ Statement I and Statement III are correct.

Given :

\(\displaystyle \int_{-π/4}^{π/4}\log(\sin x+\cos x)\:dx.\)

Formula used : 

f(x) is continuous function in [-a, a]

\(\int _{-a}^{a}f(x) dx= \int _{0}^{a}[f(x)+ f(-x)] dx\)      ----- (1)

log m + log n = log (mn)        ----- (2)

\(\int_{0}^{\pi/2}log \ sin\ x dx= \int_{0}^{\pi/2}log \ cos\ x dx= \frac{-\pi}{2}log 2\)       ----- (3)

Calculations :

Let I = \(\displaystyle \int_{-π/4}^{π/4}\log(\sin x+\cos x)\:dx.\)

Using equation (1)

⇒ \(\displaystyle \int_{0}^{π/4}\log((sin \ x+ cos\ x) + log (sin (-x) + cos (-x))dx \)

⇒ \(\displaystyle \int_{0}^{π/4}\log((sin \ x+ cos\ x) + log (cos \ x - sin \ x))dx \)

Using equation (2)

⇒ \(\displaystyle \int_{0}^{π/4}\log(cos^2x-sin^2x)dx\)

⇒ \(\displaystyle \int_{0}^{π/4}\log(cos \ 2 x)dx\)

Let, t = 2x, differentiating both sides we get

⇒ dt/2 = dx

Put value of x = 0 then t = 0 and x = π/4 then t = π/2

Limit is changed to [ 0 to π/2]

⇒ \(\displaystyle \int_{0}^{π/2}\log(cos \ t)dt\)

Using equation (3)

⇒ \(\frac{1}{2}.[-\frac{\pi}{2}\log2]\)

∴ The value of \(\displaystyle \int_{-\pi/4}^{\pi/4}\log(\sin x+\cos x)\:dx = -\frac{\pi}{4}log \ 2\)

Concept:

Integral Property:

\(\rm \displaystyle\int e^{x}\;dx = e^{x} + c\)

Calculation:

\(\rm I = \displaystyle \int\limits_{0}^{1}{\frac{e^{\tan^{-1}x}dx}{1+x^{2}}}\)

Let tan^-1 x = t

Differentiating both sides, we get

\(⇒\rm \frac{dx}{1+x^{2}} = dt\)

\(\rm I = \displaystyle\int\limits_{0}^{\frac{π}{4}} e^{t}dt\)

\(\rm ⇒ I = [{e^{t}}]_{0}^{\frac{π}{4}}\)

\(\rm \Rightarrow I = e^{\frac{\pi}{4}} - e^0\)

\(\rm \Rightarrow I = e^{\frac{\pi}{4}} - 1\)

Concept:

\(\smallint {\rm{xdx}} = \frac{{{{\rm{x}}^2}}}{2} + {\rm{c}}\)

Property of definite integrals:  \(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = - \mathop \smallint \nolimits_{\rm{b}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx\;}}\)

Calculation:

I = \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} {\cos ^2}{\rm{x}}\sin 2{\rm{xdx}}\)

\( = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} {\cos ^2}{\rm{x}}\left( {2\sin {\rm{x}}\cos {\rm{x}}} \right){\rm{dx}}\)        (∵ sin 2x = 2 sin x cos x)

\( = 2\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} {\cos ^3}{\rm{x}}\sin {\rm{xdx}}\)

Let cos x = t

⇒ -sin x dx = dt

∴ sin x dx = -dt

\( = 2\mathop \smallint \nolimits_1^0 {{\rm{t}}^3}\left( { - {\rm{dt}}} \right)\)

\( = 2\mathop \smallint \nolimits_0^1 {{\rm{t}}^3}{\rm{dt}}\)

\( = 2\left[ {\frac{{{{\rm{t}}^4}}}{4}} \right]_0^1\)

\( = {\rm{\;}}2\left( {\frac{1}{4}{\rm{\;}}-{\rm{\;}}0} \right) = \frac{1}{2}\)

Concept:

• \(\displaystyle \int_{a}^{b}f (t) \ dt = -\int_{b}^{a}f (t) \ dt\)
• Substitution method, 

             If we put x = t, then dx = dt and,

             \(\displaystyle \int_{a(x)}^{b(x)}f (x) \ dx = \int_{a(t)}^{b(t)}f (t) \ dt\)

Calculation:

Let I =\(\displaystyle \int_{-π/4}^{π/4}\log \left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)\:dx.\)

Dividing by cos x in numerator and denominator,

⇒ I = \(\displaystyle \int_{-π/4}^{π/4}\log \left(\frac{\tan x+1}{ 1-\tan x}\right)\:dx\)

⇒ I = \(\displaystyle \int_{-π/4}^{π/4}\log \left(\frac{\tan x+tan (π/4)}{ 1-\tan (π/4).tan x}\right)\:dx\)

⇒ I = \(\displaystyle \int_{-π/4}^{π/4}\log \left({\tan ({π \over 4} + x)}\right)\:dx\)

Put \({π \over 4}+x = t\), 

⇒ dx = dt and 

when x = - π/4 then t = 0

when x = π/4 then t = π/2

⇒ I = \(\displaystyle \int_{0}^{\pi/2}\log \left({\tan (t)}\right)\:(dt)\)

⇒ I = \(\displaystyle \int_{0}^{π/2}\log \left({\tan \ t}\right)\:dt\)

⇒ I = 0 (using the formula in comprehension)

∴ The correct answer is an option (3).

Given, I = \(\int\limits_0^{\frac{π }{2}} {\sin x\,\sin 2x} \ dx\)

⇒ I = \(\int\limits_0^{\frac{π }{2}} {\sin x\,2\sin x \cos x} \ dx\)  { ∵ sin 2x = 2 sin x cos x}

⇒ I = \(2\int\limits_0^{\frac{π }{2}} {\sin^2 x\, \cos x} \ dx\) 

Put sin x = t → cos x dx  = dt

when x = 0, t = sin 0 = 0 and when x = π/2, t = sin π/2 = 1

⇒ I = \(2\int\limits_0^1{t^2 }dt\) = \({2t^3 \over 3} |_0^1\)

⇒ I = 2/3

∴ The correct answer is option (4).

Concept:

• sin 2x = 2 sin x cos x
• sin²x + cos²x = 1
• \(\int {dx \over a^2 -x^2} = {1 \over 2a} \log |{a +x \over a-x}| + C\)

Calculation:

Put sin x - cos x = t ⇒ (cos x + sin x) dx = dt

⇒ (sin x - cos x)² = t²

⇒ sin2x + cos2x - 2sinx cos x = t2

⇒ 1 - sin 2x = t2

⇒ sin 2x = 1 - t2

And when x = 0 → t = sin 0 - cos 0 = - 1

when x = π/4 → t = sin (π/4) - cos (π/4) = 0 

⇒ \(I = \int\limits_{-1}^{0} {\frac{dt}{{9+16(1 - t^2)}}} \, \)

⇒ \(I = \int\limits_{-1}^{0} {\frac{dt}{{25 - 16t^2}}} \, \)

⇒ \(I = {1 \over 16} \int\limits_{-1}^{0} {\frac{dt}{{({5 \over4})^2 -t^2}}} \, \)

⇒ \(I = {1 \over 16} {4 \over 2(5)}\log \Big|{{5 \over4}+ t \over {5 \over 4} - t}\Big| _{-1}^0\)

⇒ \(I = {1 \over 40} \log \Big|{5+ 4t \over5 - 4t}\Big|_{-1}^0\)

⇒ \(I = {1 \over 40} \Big|\log ({5+ 0 \over5 - 0}) - \log ({5 -4 \over5 +4})\Big|\)

⇒ \(I = {1 \over 40} \Big|\log ({5\over5 }) - \log ({1 \over9})\Big|\)

⇒ \(I = {1 \over 40} ( \log 1 +\log 9)\)

⇒ \(I = {1 \over 40} ( \log 9)\)

⇒ \(I = {2 \over 40} \log 3 \)

⇒ \(I = {1 \over 20} \log 3 \)

∴ The correct answer is option (1).