Để tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiabg2 % da9maapehabaGaamiEaiGacohacaGGPbGaaiOBaiaaykW7caaIXaGa % aGOmaiaadIhacaqGKbGaamiEaaWcbaGaaGimaaqaaiabec8aWbqdcq % GHRiI8aaaa!465F! H = \int\limits_0^\pi {x\sin \,12x{\rm{d}}x} \) bằng phương pháp tích phân từng phần ta đặt u = x và dv = sin12xdx. Tìm du và tính H.
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpcaWG4baabaGaaeizaabaaaaaaaaapeGaamOD % aiabg2da9iGacohacaGGPbGaaiOBaiaaykW7caaIXaGaaGOmaiaadI % hacaqGKbGaamiEaaaapaGaay5Eaaaaaa!45E4! \left\{ \begin{array}{l} u = x\\ {\rm{d}}v = \sin \,12x{\rm{d}}x \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aai % qaaqaabeqaaabaaaaaaaaapeGaaeizaiaadwhacqGH9aqpcaqGKbGa % amiEaaqaaiaadAhacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaaca % aIXaGaaGOmaaaaciGGJbGaai4BaiaacohacaaIXaGaaGOmaiaadIha % aaWdaiaawUhaaaaa!48E3! \Rightarrow \left\{ \begin{array}{l} {\rm{d}}u = {\rm{d}}x\\ v = - \frac{1}{{12}}\cos 12x \end{array} \right.\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiabg2 % da9iabgkHiTmaalaaabaGaaGymaaqaaiaaigdacaaIYaaaaiaadIha % ciGGJbGaai4BaiaacohacaaIXaGaaGOmaiaadIhadaabbaabaeqaba % GaeqiWdahabaGaaGimaaaacaGLhWoacqGHRaWkdaWdXbqaamaalaaa % baGaaGymaaqaaiaaigdacaaIYaaaaiaabogacaqGVbGaae4Caiaabg % dacaqGYaGaamiEaiaabsgacaWG4baaleaacaaIWaaabaGaeqiWdaha % niabgUIiYdaaaa!5467! H = - \frac{1}{{12}}x\cos 12x\left| \begin{array}{l} \pi \\ 0 \end{array} \right. + \int\limits_0^\pi {\frac{1}{{12}}{\rm{cos12}}x{\rm{d}}x} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaey % OeI0YaaSaaaeaacaaIXaaabaGaaGymaiaaikdaaaGaamiEaiGacoga % caGGVbGaai4CaiaaigdacaaIYaGaamiEamaaeeaaeaqabeaacqaHap % aCaeaacaaIWaaaaiaawEa7aiabgUcaRmaalaaabaGaaGymaaqaaiaa % igdacaaIYaWaaWbaaSqabeaacaaIYaaaaaaakiGacohacaGGPbGaai % OBaiaaykW7caaIXaGaaGOmaiaadIhadaabbaabaeqabaGaeqiWdaha % baGaaGimaaaacaGLhWoacqGH9aqpcqGHsisldaWcaaqaaiabec8aWb % qaaiaaigdacaaIYaaaaaaa!58B5! = - \frac{1}{{12}}x\cos 12x\left| \begin{array}{l} \pi \\ 0 \end{array} \right. + \frac{1}{{{{12}^2}}}\sin \,12x\left| \begin{array}{l} \pi \\ 0 \end{array} \right. = - \frac{\pi }{{12}}\)