Интеграл

[Hitp]

Помогите разобраться с интегралом $% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaci % GGJbGaai4BaiaacohacaGGOaGaaGymaiabgkHiTiaaikdacaWG0bGa % eyOeI0IaamyAaiaadshacaGGPaGaamizaiaacIcacaaIXaGaey4kaS % IaamyAaiaadshacaGGPaaaleaacaaIWaaabaGaaGymaaqdcqGHRiI8 % aaaa!49F5! \[\int\limits_0^1 {\cos (1 - 2t - it)d(1 + it)} \]$ Могу ли я написать, что $% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacI % cacaaIXaGaey4kaSIaamyAaiaadshacaGGPaaaaa!3BBA! \[{d(1 + it)}\]$ = dt?

[Null]

$d(1 + it) = (1+it)'dt$ - еще на производную нужно умножить.

(Оффтоп)

Что-то у вас формулы странные

[Hitp]

тогда получится$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyAaiaads % gacaWG0baaaa!38C4! \[idt\]$?

[Everest]

Да, верно.

$$\int^1_0 cos(1-2t-it) d(1+it)=i \int^1_0 cos(1-2t-it) dt=-\frac{i}{2+i}\int^1_0 cos(1-(2+i)t) d(1-(2+i)t)=-\frac{i}{2+i} (sin(1-2-i)-sin(1))=$$ $$=-\frac{i}{2+i} (sin(1-2-i)-sin(1))=\frac{i}{2+i} (sin(1+i)+sin(1))$$