February-March 2023 Paper 2 Pure Mathematics 2 - Edubirdie

University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

172

Pages:

2
Author:

Shahbaz A.

Cambridge International AS and A Level 9709/22 February/March 2023 Paper 2 Pure Mathematics 2 shahbaz ahmed July 2024 1 Find the exact value of R 1π 2 0 Solution R 21 π R 1π 2 tan2 12 x dx =2 02 tan2 12 x dx 0 Since R R tan2 θ dθ = (sec2 θ − 1) dθ R R R tan2 θ dθ = sec2 θ dθ − dθ R tan2 θ dθ = tan θ − θ + C ′ Putting θ = 12 x dθ = d( 12 x) dθ = 12 dx R tan2 21 x ( 12 dx) = tan 12 x − 12 x + C ′ R 1 tan2 21 x dx = tan 12 x − 12 x + C ′ 2 R tan2 12 x dx = 2 tan 12 x − 2( 12 )x + 2C ′ R tan2 21 x dx = 2 tan 12 x − x + 2C ′ Putting 2C ′ = C R tan2 12 x dx = 2 tan 12 x − x + C Applying the limit 1 2 tan2 12 x dx 1π tan2 12 x dx = 2 tan 21 x − x + C 02 R 12 π tan2 12 x dx = 2 tan( 12 × 12 π) − 12 π + C − 2 tan 12 (0) + (0) − C 0 R 21 π tan2 12 x dx = 2 − 12 π 0 R 1π 2 02 tan2 12 x dx = 2 2 − 12 π R 1π 2 02 tan2 12 x dx = 4 − π R 12 π 2 tan2 12 x dx = 4 − π 0 R 1 2π 0 2

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