Question 3 - Case Based Questions (MCQ) - Chapter 7 Class 12 Integrals (Term 2)
Last updated at March 16, 2023 by Teachoo
The given integral β« f(x) dx can be transformed into another form by changing the independent variable x to t by substituting x=g(t)
Β
Consider
Β I = β« f(x) dx
Put
Β Β x = g(t)
so that
dx/dt = gβ²(t)
We writeΒ
dx = gβ² (t)dt
ThusΒ
I = β« f(x) dx= β« f(g(t))Β gβ²(t) dt
Β
This change of variable formula is one of the important tools available to us in the name of integration by substitution.
Question 1
β« sinβ‘ (tan
-1
x)/(1 + x
2
) dx is equal to:
(a) -sinβ‘(tan
-1
β‘x + CΒ
(b) -cosβ‘(tan
-1
β‘x + C
(c) tanβ‘x + CΒ
(d) None of these
Question 2
β« tanβ‘xΒ dx is equal to:
(a) secβ‘x + C Β Β
(b) cotβ‘x + C Β
(c) logβ‘|x| + C Β Β
(d) None of these
Question 3
β« 2x/1 + x
2
Β dx is equal to:
(a) 1 + x
2
+ C Β
(b) logβ‘ |1 + x
2
| + C
(c) log β‘|2/1 + x
2
| + CΒ
(d) None of these
Question 4
β« sinβ‘ (ax+b)Β cosβ‘ (ax+b)Β dx is equal to:
(a) cos
2
Β x (ax + b) + C Β
(b) sin
2
β‘(ax + b)
(c) (-1)/4aΒ cosβ‘2Β (ax + b) + CΒ
(d) None of these
This question is inspired from
Ex 7.2, 5 - Chapter 7 Class 12
- Integrals
Β
Question 5
β« 1/x + x logβ‘xΒ dx is equal to:
(a) |1 + logβ‘x | + CΒ
(b) logβ‘|1 + logβ‘x| + C
(c) logβ‘x + CΒ
(d) None of these
This question is inspired from
Ex 7.2, 3 - Chapter 7 Class 12
- Integrals
Get live Maths 1-on-1 Classs - Class 6 to 12
Transcript
Question 3 The given integral β«1βγπ(π₯)γ ππ₯ can be transformed into another form by changing the independent variable π₯ to π‘ by substituting π₯=π(π‘) Consider I = β«1βγπ(π₯)γ ππ₯ Put π₯ = π(π‘) so that ππ₯/ππ‘=πβ²(π‘) We write ππ₯=π^β² (π‘)ππ‘ Thus I = β«1βγπ(π₯)γ ππ₯= β«1βγπ(π(π‘)) π^β² (π‘) ππ‘γ This change of variable formula is one of the important tools available to us in the name of integration by substitution.
For example: β«1βγ2π₯ sinβ‘γ(π₯^2+1)γ ππ₯γ Put π₯^2+1=π‘ 2π₯ππ₯=ππ‘ Thus, β«1βγsinβ‘γ(π‘) ππ‘γ=βcosβ‘(π‘)+πΆγ = βcosβ‘γ(π₯^2+1)+πΆγ Based on the above information, answer any four of the following questions.
Question 1 β«1βγsinβ‘γ(tan^(β1)β‘π₯)γ/(1 + π₯^2 ) ππ₯γ is equal to: (a) βsinβ‘(tan^(β1)β‘π₯ )+πΆ (b) βcosβ‘(tan^(β1)β‘π₯ )+πΆ (c) tanβ‘π₯+πΆ (d) None of these
β«1βγsinβ‘γ(tan^(β1)β‘π₯)γ/(1 + π₯^2 ) ππ₯γ
Let γπππ§γ^(βπ)β‘π=π
Differentiating both sides π€.π.π‘.π₯
π/(π + π^π )= π
π/π
π
ππ₯/(1 + π₯^2 )=ππ‘
Now,
Integrating the function
β«1βγπππβ‘γ(γπππγ^(βπ)β‘π)γ/(π + π^π ) π
πγ
Putting tan^(β1)β‘π₯=π‘ & ππ₯/(1 + π₯^2 )=ππ‘
= β«1βγπ¬π’π§β‘π . π
πγ
= βcosβ‘π‘+πΆ
Putting back t = tan^(β1)β‘π₯
= βππ¨π¬β‘γ(πππ^(βπ) π)γ+πͺ
So, the correct answer is (b)
Question 2 β«1βγtanβ‘π₯ ππ₯γ is equal to: (a) secβ‘π₯+πΆ (b) cotβ‘π₯+πΆ (c) logβ‘γ|π₯|γ+πΆ (d) None of these
β«1βγtanβ‘π₯ ππ₯γ=β«1βγsinβ‘π₯/cosβ‘π₯ ππ₯γ
Let πππβ‘π=π
Differentiating both sides π€.π.π‘.π₯
βπππ π= π
π/π
π
sin x ππ₯=βππ‘
Now,
β«1β(πππ π)/(πππ π) π
π
Putting πππ β‘π₯=π‘ & sinβ‘π₯ ππ₯=ππ‘
= β«1βγπ/π . (βπ
π)γ
= ββ«1βπ
π/π
= βlogβ‘γ|π‘|γ+πΆ
Putting back t = cosβ‘π₯
= βπ₯π¨π |πππ π|+πͺ
= loπ |πππ π₯|^(β1)+πΆ
= loπ 1/(|πππ π₯|)+πΆ
= π₯π¨π |πππ π|+πͺ
So, the correct answer is (d)
Question 3 β«1βγ2π₯/(1 + π₯^2 ) ππ₯γ is equal to: (a) 1+π₯^2+πΆ (b) logβ‘γ|1+π₯^2 |+πΆγ (c) logβ‘|2/(1 + π₯^2 )|+πΆ (d) None of these
β«1βγ2π₯/(1 + π₯^2 ) ππ₯γ
Let π+π^π=π
Differentiating both sides π€.π.π‘.π₯
ππ= π
π/π
π
2π₯ ππ₯=ππ‘
Now,
Integrating the function
β«1βγππ/(π + π^π ) π
πγ
Putting 1+π₯^2=π‘ & 2π₯ππ₯=ππ‘
= β«1βγπ/π . π
πγ
= logβ‘γ|π‘|γ+πΆ
Putting back t = 1+π₯^2
= π₯ππ|π+π^π |+πͺ
So, the correct answer is (b)
Question 4 β«1βγsinβ‘γ(ππ₯+π)γ cosβ‘γ(ππ₯+π)γ ππ₯γ is equal to: (a) cos^2 π₯ (ππ₯+π)+πΆ (b) sin^2β‘γ(ππ₯+π)γ (c) (β1)/4π cosβ‘2 (ππ₯+π)+πΆ (d) None of these
Taking the given function
πππβ‘(ππ + π) πππβ‘(ππ + π)
= 1/2 sinβ‘(2(ππ₯+π))
= π/π πππβ‘(πππ+ππ)
Let (πππ+ππ)=π
Differentiating both sides π€.π.π‘.π₯
2π+0=ππ‘/ππ₯
2π=ππ‘/ππ₯
2π.ππ₯=ππ‘
π
π=π
π/ππ
Integrating the function
β«1βγ" " sinβ‘(ππ₯+π) cosβ‘(ππ₯+π)" " γ.ππ₯
= 1/2 β«1βγsinβ‘(2ππ₯+2π)γ .ππ₯
Putting π‘=2ππ₯+2π & ππ₯=ππ‘/2π
= π/π β«1βγπππ πγ .π
π/ππ
= 1/4π β«1βγsinβ‘(π‘)γ .ππ‘
= 1/4π [βcosβ‘π‘+πΆ1]
= β 1/4π . cosβ‘π‘ + πΆ1/4π
= β π/ππ . πππβ‘π + πͺ
Putting back π‘=2ππ₯+2π
= β 1/4π . cosβ‘γ(2ππ₯+2π)γ + πΆ
= β π/ππ . πππβ‘γπ(ππ+π)γ+πͺ
So, the correct answer is (c)
Question 5 β«1βγ1/(π₯ + π₯ logβ‘π₯ ) ππ₯γ is equal to: (a) |1+logβ‘π₯ |+πΆ (b) logβ‘γ|1+logβ‘π₯ |+πΆγ (c) logβ‘π₯+πΆ (d) None of these
β«1βγ1/(π₯ + π₯ logβ‘π₯ ) ππ₯γ = β«1βγ1/(π₯(1 + logβ‘π₯)) ππ₯γ
Let 1 + log π= π
Differentiate π€.π.π‘.π₯
0 + ππ‘/ππ₯= 1/π₯
0 + ππ‘/ππ₯= 1/π₯
π
π.π=π
π
Now,
β«1β1/(π₯ + π₯ logβ‘π₯ ) .ππ₯ =β«1β1/(π₯ (1 + logβ‘π₯ ) ) .ππ₯" "
Putting 1+logβ‘π₯ & ππ₯=π₯ ππ‘
= β«1β1/(π₯(π‘)) ππ‘.π₯
= β«1β1/π‘ ππ‘
= πππ|π‘|+πΆ
Putting back π‘ =1+πππβ‘π₯
= πππ|π+π₯π¨π β‘π |+πͺ
So, the correct answer is (b)
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