Tích phân \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiaadIhaaeaacaaIXaGaey4kaSIaci4yaiaac+gacaGGZbGa % aGOmaiaadIhaaaGaaeizaiaadIhacqGH9aqpcaWGHbGaeqiWdaNaey % 4kaSIaamOyaiGacYgacaGGUbGaaGOmaaWcbaGaaGimaaqaamaalaaa % baGaeqiWdahabaGaaGinaaaaa0Gaey4kIipaaaa!4CCE! \int\limits_0^{\frac{\pi }{4}} {\frac{x}{{1 + \cos 2x}}{\rm{d}}x = a\pi + b\ln 2} \), với a, b là các số thực . tính 16a - 8b.
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Lời giải:
Báo saiĐặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpcaWG4baabaGaaeizaiaadAhacqGH9aqpdaWc % aaqaaiaabsgacaWG4baabaGaaGymaiabgUcaRiGacogacaGGVbGaai % 4CaiaaikdacaWG4baaaaaacaGL7baacqGHshI3daGabaabaeqabaGa % aeizaiaadwhacqGH9aqpcaqGKbGaamiEaaqaaiaadAhacqGH9aqpda % WcaaqaaiaaigdaaeaacaaIYaaaaiGacshacaGGHbGaaiOBaiaadIha % aaGaay5Eaaaaaa!54B6! \left\{ \begin{array}{l} u = x\\ {\rm{d}}v = \frac{{{\rm{d}}x}}{{1 + \cos 2x}} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {\rm{d}}u = {\rm{d}}x\\ v = \frac{1}{2}\tan x \end{array} \right.\)
Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maalaaabaGaaGymaaqaaiaaikdaaaGaamiEaiGacshacaGGHbGa % aiOBaiaadIhadaabbaabaeqabaWaaSaaaeaacqaHapaCaeaacaaI0a % aaaaqaaiaaicdaaaGaay5bSdGaeyOeI0YaaSaaaeaacaaIXaaabaGa % aGOmaaaadaWdXaqaaiGacshacaGGHbGaaiOBaiaadIhacaqGKbGaam % iEaaWcbaGaaGimaaqaamaalaaabaGaeqiWdahabaGaaGinaaaaa0Ga % ey4kIipakiabg2da9maalaaabaGaeqiWdahabaGaaGioaaaacqGHRa % WkdaWcaaqaaiaaigdaaeaacaaIYaaaaiGacYgacaGGUbWaaqWaaeaa % ciGGJbGaai4BaiaacohacaWG4baacaGLhWUaayjcSdWaaqqaaqaabe % qaamaalaaabaGaeqiWdahabaGaaGinaaaaaeaacaaIWaaaaiaawEa7 % aiabg2da9maalaaabaGaeqiWdahabaGaaGioaaaacqGHRaWkdaWcaa % qaaiaaigdaaeaacaaIYaaaaiGacYgacaGGUbWaaSaaaeaacaaIXaaa % baWaaOaaaeaacaaIYaaaleqaaaaakiabg2da9maalaaabaGaeqiWda % habaGaaGioaaaacqGHsisldaWcaaqaaiaaigdaaeaacaaI0aaaaiGa % cYgacaGGUbGaaGOmaiabgkDiElaadggacqGH9aqpdaWcaaqaaiaaig % daaeaacaaI4aaaaiaacYcacaWGIbGaeyypa0JaeyOeI0YaaSaaaeaa % caaIXaaabaGaaGinaaaaaaa!816E! I = \frac{1}{2}x\tan x\left| \begin{array}{l} \frac{\pi }{4}\\ 0 \end{array} \right. - \frac{1}{2}\int_0^{\frac{\pi }{4}} {\tan x{\rm{d}}x} = \frac{\pi }{8} + \frac{1}{2}\ln \left| {\cos x} \right|\left| \begin{array}{l} \frac{\pi }{4}\\ 0 \end{array} \right. = \frac{\pi }{8} + \frac{1}{2}\ln \frac{1}{{\sqrt 2 }} = \frac{\pi }{8} - \frac{1}{4}\ln 2 \Rightarrow a = \frac{1}{8},b = - \frac{1}{4}\)
vậy 16a - 8b = 4
