Case Based Questions (MCQ)

Chapter 7 Class 12 Integrals
Serial order wise

## (d) None of these

This question is inspired from Ex 7.2, 5 - Chapter 7 Class 12 - Integrals

## (d) None of these

This question is inspired from Ex 7.2, 3 - Chapter 7 Class 12 - Integrals

### Transcript

Question 3 The given integral β«1βγπ(π₯)γ ππ₯ can be transformed into another form by changing the independent variable π₯ to π‘ by substituting π₯=π(π‘) Consider I = β«1βγπ(π₯)γ ππ₯ Put π₯ = π(π‘) so that ππ₯/ππ‘=πβ²(π‘) We write ππ₯=π^β² (π‘)ππ‘ Thus I = β«1βγπ(π₯)γ ππ₯= β«1βγπ(π(π‘)) π^β² (π‘) ππ‘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«1βγ2π₯ sinβ‘γ(π₯^2+1)γ ππ₯γ Put π₯^2+1=π‘ 2π₯ππ₯=ππ‘ Thus, β«1βγsinβ‘γ(π‘) ππ‘γ=βcosβ‘(π‘)+πΆγ = βcosβ‘γ(π₯^2+1)+πΆγ Based on the above information, answer any four of the following questions. Question 1 β«1βγsinβ‘γ(tan^(β1)β‘π₯)γ/(1 + π₯^2 ) ππ₯γ is equal to: (a) βsinβ‘(tan^(β1)β‘π₯ )+πΆ (b) βcosβ‘(tan^(β1)β‘π₯ )+πΆ (c) tanβ‘π₯+πΆ (d) None of these β«1βγsinβ‘γ(tan^(β1)β‘π₯)γ/(1 + π₯^2 ) ππ₯γ Let γπ­ππ§γ^(βπ)β‘π=π Differentiating both sides π€.π.π‘.π₯ π/(π + π^π )= ππ/ππ ππ₯/(1 + π₯^2 )=ππ‘ Now, Integrating the function β«1βγπππβ‘γ(γπππγ^(βπ)β‘π)γ/(π + π^π ) ππγ Putting tan^(β1)β‘π₯=π‘ & ππ₯/(1 + π₯^2 )=ππ‘ = β«1βγπ¬π’π§β‘π . ππγ = βcosβ‘π‘+πΆ Putting back t = tan^(β1)β‘π₯ = βππ¨π¬β‘γ(πππ^(βπ) π)γ+πͺ So, the correct answer is (b) Question 2 β«1βγtanβ‘π₯ ππ₯γ is equal to: (a) secβ‘π₯+πΆ (b) cotβ‘π₯+πΆ (c) logβ‘γ|π₯|γ+πΆ (d) None of these β«1βγtanβ‘π₯ ππ₯γ=β«1βγsinβ‘π₯/cosβ‘π₯ ππ₯γ Let πππβ‘π=π Differentiating both sides π€.π.π‘.π₯ βπππ π= ππ/ππ sin x ππ₯=βππ‘ Now, β«1β(πππ π)/(πππ π) ππ Putting πππ β‘π₯=π‘ & sinβ‘π₯ ππ₯=ππ‘ = β«1βγπ/π . (βππ)γ = ββ«1βππ/π = βlogβ‘γ|π‘|γ+πΆ Putting back t = cosβ‘π₯ = βπ₯π¨π |πππ π|+πͺ = loπ |πππ  π₯|^(β1)+πΆ = loπ 1/(|πππ  π₯|)+πΆ = π₯π¨π |πππ π|+πͺ So, the correct answer is (d) Question 3 β«1βγ2π₯/(1 + π₯^2 ) ππ₯γ is equal to: (a) 1+π₯^2+πΆ (b) logβ‘γ|1+π₯^2 |+πΆγ (c) logβ‘|2/(1 + π₯^2 )|+πΆ (d) None of these β«1βγ2π₯/(1 + π₯^2 ) ππ₯γ Let π+π^π=π Differentiating both sides π€.π.π‘.π₯ ππ= ππ/ππ 2π₯ ππ₯=ππ‘ Now, Integrating the function β«1βγππ/(π + π^π ) ππγ Putting 1+π₯^2=π‘ & 2π₯ππ₯=ππ‘ = β«1βγπ/π . ππγ = logβ‘γ|π‘|γ+πΆ Putting back t = 1+π₯^2 = π₯ππ|π+π^π |+πͺ So, the correct answer is (b) Question 4 β«1βγsinβ‘γ(ππ₯+π)γ cosβ‘γ(ππ₯+π)γ ππ₯γ is equal to: (a) cos^2 π₯ (ππ₯+π)+πΆ (b) sin^2β‘γ(ππ₯+π)γ (c) (β1)/4π cosβ‘2 (ππ₯+π)+πΆ (d) None of these Taking the given function πππβ‘(ππ + π) πππβ‘(ππ + π) = 1/2 sinβ‘(2(ππ₯+π)) = π/π πππβ‘(πππ+ππ) Let (πππ+ππ)=π Differentiating both sides π€.π.π‘.π₯ 2π+0=ππ‘/ππ₯ 2π=ππ‘/ππ₯ 2π.ππ₯=ππ‘ ππ=ππ/ππ Integrating the function β«1βγ" " sinβ‘(ππ₯+π) cosβ‘(ππ₯+π)" " γ.ππ₯ = 1/2 β«1βγsinβ‘(2ππ₯+2π)γ .ππ₯ Putting π‘=2ππ₯+2π & ππ₯=ππ‘/2π = π/π β«1βγπππ πγ .ππ/ππ = 1/4π β«1βγsinβ‘(π‘)γ .ππ‘ = 1/4π [βcosβ‘π‘+πΆ1] = β 1/4π . cosβ‘π‘ + πΆ1/4π = β π/ππ . πππβ‘π + πͺ Putting back π‘=2ππ₯+2π = β 1/4π . cosβ‘γ(2ππ₯+2π)γ + πΆ = β π/ππ . πππβ‘γπ(ππ+π)γ+πͺ So, the correct answer is (c) Question 5 β«1βγ1/(π₯ + π₯ logβ‘π₯ ) ππ₯γ is equal to: (a) |1+logβ‘π₯ |+πΆ (b) logβ‘γ|1+logβ‘π₯ |+πΆγ (c) logβ‘π₯+πΆ (d) None of these β«1βγ1/(π₯ + π₯ logβ‘π₯ ) ππ₯γ = β«1βγ1/(π₯(1 + logβ‘π₯)) ππ₯γ Let 1 + log π= π Differentiate π€.π.π‘.π₯ 0 + ππ‘/ππ₯= 1/π₯ 0 + ππ‘/ππ₯= 1/π₯ ππ.π=ππ Now, β«1β1/(π₯ + π₯ logβ‘π₯ ) .ππ₯ =β«1β1/(π₯ (1 + logβ‘π₯ ) ) .ππ₯" " Putting 1+logβ‘π₯ & ππ₯=π₯ ππ‘ = β«1β1/(π₯(π‘)) ππ‘.π₯ = β«1β1/π‘ ππ‘ = πππ|π‘|+πΆ Putting back π‘ =1+πππβ‘π₯ = πππ|π+π₯π¨π β‘π |+πͺ So, the correct answer is (b)