CBSE Class 12 Sample Paper for 2026 Boards
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CBSE Class 12 Sample Paper for 2026 Boards
Last updated at Sept. 2, 2025 by Teachoo
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Question 33 (A) Evaluate: β«_0^1β(log (1 + π₯))/(1 + π₯^2 ) ππ₯We know that (π(tan^(β1)β‘π₯))/ππ₯=1/(1 + π₯^2 ) Since we have 1/(1 + π₯^2 ), we put x = tan π Thus, ππ₯=sec^2β‘π ππ Change the limits of integration: When π=π,tanβ‘π=0βΉπ½=π When π=π,tanβ‘π=1βΉπ½=π /π The denominator becomes 1+π₯^2=1+tan^2β‘π=γπππγ^πβ‘π½. Now, πΌ=β«130_0^1ββ (logβ‘(1 + π₯))/(1 + π₯^2 ) ππ₯ After Substitution πΌ=β«130_0^(π/4)βββ(logβ‘(1 + tanβ‘π))/(sec^2β‘π) sec^2β‘πππ π°=β«130_π^(π /π)βββπ₯π¨π β‘(π+πππ§β‘π½)π π½β‘( ) Now, we know that β«130_0^πββ π(π₯)ππ₯=β«130_0^πββ π(πβπ₯)ππ₯ Applying this property to Equation 1, with π=π/4 and the variable being π : πΌ=β«130_0^(π/4)ββ logβ‘(1+tanβ‘(π/4βπ))ππ Using π‘ππβ‘(π΄βπ΅)=(π‘ππβ‘π΄βπ‘ππβ‘π΅)/(1+π‘ππβ‘π΄π‘ππβ‘π΅) : πΌ=β«130_0^(π/4)ββ logβ‘(1+(tanβ‘(π/4)β tanβ‘π)/(1 + tanβ‘(π/4)tanβ‘π))ππ πΌ=β«130_0^(π/4)ββ logβ‘(1+(1 β tanβ‘π)/(1 + tanβ‘π))ππ πΌ=β«130_0^(π/4)ββ logβ‘(((1 + tanβ‘π )+ (1 βtanβ‘π ))/(1 +tanβ‘π ))ππ πΌ=β«130_0^(π/4)ββ log(2/(1+tanβ‘π))ππ Using the logarithm property logβ‘(π/π)=logβ‘πβlogβ‘π π°=β«130_π^(π /π)ββ [πππβ‘πβπππβ‘(π+πππβ‘π½ ) ]π π½β‘( ) Now, Adding (1) and (2) πΌ+πΌ=β«130_0^(π/4)βββlogβ‘(1+tanβ‘π)ππ+β«130_0^(π/4)βββ|logβ‘2βlogβ‘(1+tanβ‘π)|ππ 2πΌ=β«130_0^(π/4)βββ|logβ‘(1+tanβ‘π)+logβ‘2βlogβ‘(1+tanβ‘π)|ππ ππ°=β«130_π^(π /π)ββ πππβ‘π π π½ Since logβ‘2 is a constant: 2πΌ=logβ‘2 β«130_0^(π/4)βββ1ππ 2πΌ=log 2 Γ |π|_0^(π/4) 2πΌ=logβ‘γ2 Γγ (π/4β0) 2πΌ=π/4 log 2 π°=π /π πππβ‘π
