[Sample Paper Class 12] Evaluate: ∫0^1 (log (1+𝑥))/(1+𝑥^2) 𝑑𝑥

Question 33 (A) Evaluate: ∫_0^1 (log (1 + 𝑥))/(1 + 𝑥^2 ) 𝑑𝑥We know that (𝑑(tan^(−1)⁡𝑥))/𝑑𝑥=1/(1 + 𝑥^2 ) Since we have 1/(1 + 𝑥^2 ), we put x = tan 𝜃 Thus, 𝑑𝑥=sec^2⁡𝜃 𝑑𝜃 Change the limits of integration: When 𝒙=𝟎,tan⁡𝜃=0⟹𝜽=𝟎 When 𝒙=𝟏,tan⁡𝜃=1⟹𝜽=𝝅/𝟒 The denominator becomes 1+𝑥^2=1+tan^2⁡𝜃=〖𝒔𝒆𝒄〗^𝟐⁡𝜽. Now, 𝐼=∫130_0^1▒ (log⁡(1 + 𝑥))/(1 + 𝑥^2 ) 𝑑𝑥 After Substitution 𝐼=∫130_0^(𝜋/4)▒ (log⁡(1 + tan⁡𝜃))/(sec^2⁡𝜃) sec^2⁡𝜃𝑑𝜃 𝑰=∫130_𝟎^(𝝅/𝟒)▒ 𝐥𝐨𝐠⁡(𝟏+𝐭𝐚𝐧⁡𝜽)𝒅𝜽□( ) Now, we know that ∫130_0^𝑎▒ 𝑓(𝑥)𝑑𝑥=∫130_0^𝑎▒ 𝑓(𝑎−𝑥)𝑑𝑥 Applying this property to Equation 1, with 𝑎=𝜋/4 and the variable being 𝜃 : 𝐼=∫130_0^(𝜋/4)▒ log⁡(1+tan⁡(𝜋/4−𝜃))𝑑𝜃 Using 𝑡𝑎𝑛⁡(𝐴−𝐵)=(𝑡𝑎𝑛⁡𝐴−𝑡𝑎𝑛⁡𝐵)/(1+𝑡𝑎𝑛⁡𝐴𝑡𝑎𝑛⁡𝐵) : 𝐼=∫130_0^(𝜋/4)▒ log⁡(1+(tan⁡(𝜋/4)− tan⁡𝜃)/(1 + tan⁡(𝜋/4)tan⁡𝜃))𝑑𝜃 𝐼=∫130_0^(𝜋/4)▒ log⁡(1+(1 − tan⁡𝜃)/(1 + tan⁡𝜃))𝑑𝜃 𝐼=∫130_0^(𝜋/4)▒ log⁡(((1 + tan⁡𝜃 )+ (1 −tan⁡𝜃 ))/(1 +tan⁡𝜃 ))𝑑𝜃 𝐼=∫130_0^(𝜋/4)▒ log(2/(1+tan⁡𝜃))𝑑𝜃 Using the logarithm property log⁡(𝑎/𝑏)=log⁡𝑎−log⁡𝑏 𝑰=∫130_𝟎^(𝝅/𝟒)▒ [𝒍𝒐𝒈⁡𝟐−𝒍𝒐𝒈⁡(𝟏+𝒕𝒂𝒏⁡𝜽 ) ]𝒅𝜽□( ) Now, Adding (1) and (2) 𝐼+𝐼=∫130_0^(𝜋/4)▒ log⁡(1+tan⁡𝜃)𝑑𝜃+∫130_0^(𝜋/4)▒ |log⁡2−log⁡(1+tan⁡𝜃)|𝑑𝜃 2𝐼=∫130_0^(𝜋/4)▒ |log⁡(1+tan⁡𝜃)+log⁡2−log⁡(1+tan⁡𝜃)|𝑑𝜃 𝟐𝑰=∫130_𝟎^(𝝅/𝟒)▒ 𝒍𝒐𝒈⁡𝟐 𝒅𝜽 Since log⁡2 is a constant: 2𝐼=log⁡2 ∫130_0^(𝜋/4)▒ 1𝑑𝜃 2𝐼=log 2 × |𝜃|_0^(𝜋/4) 2𝐼=log⁡〖2 ×〗 (𝜋/4−0) 2𝐼=𝜋/4 log 2 𝑰=𝝅/𝟖 𝒍𝒐𝒈⁡𝟐

Question 33 (A) - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Davneet Singh

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular