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Finding definite integral in advanced mathematics
x^2 - 2ax = (x-a)^2 - a^2

Let x-a = asecu, then x = a (1+secu), dx = asecutanudu.

I =∫& lt; -π,0 & gta( 1+secu)atanu asecutanu du

= a^3∫& lt; -π, 0> secu (1+secu) (tanu) 2 du

= a^3∫& lt; -π, 0 > Du

= a^3∫& lt; -π, 0 > Du

= a^3∫& lt; -π, 0 > [(SECU) 4+(SECU) 3-(SECU) 2-SECU) Du

= a^3[∫& lt; -π,0 & gt(secu)^4du+∫& lt; -π,0 & gt(secu)^3du - 0 - 0)

= a^3[∫& lt; -π,0 & gt(secu)^4du+∫& lt; -π,0 & gt(secu)^3du)

I 1 =∫& lt; -π,0 & gt(secu)^4du =∫& lt; -π,0 & gt[ 1+ (tanu)^2]dtanu = 0

I2 =∫& lt; -π,0 & gt(secu)^3du =∫& lt; -π,0 & gtsecudtanu

=[secutanu]& lt; -π,0 & gt-∫& lt; -π,0 & gtsecu(tanu)^2du

= 0-∫& lt; -π,0 & gt(secu)^3du+∫& lt; -π, 0> Sekudu

= -I2 + 0

I2 = 0 is obtained.

I = 0