Вопрос
ct all expressions matching the value of indefinite Iral int tan(t)/(2)dt 2cos^2t+C (2)/(cos^2)(t)/(2)+C -2ln(cos(t)/(2))+c -ln(cos(t)/(2))+c
Решения
4.2256 голоса
Харлам
элита · Репетитор 8 летЭкспертная проверка
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The correct answer is:
\frac {2}{cos^{2}\frac {t}{2}}+C
Explanation:
The integral \int \tan\frac {t}{2}dt
\int \tan\frac {t}{2}dt = \int \tan u \cdot 2du = 2\int \tan u du
Now, we can use the identity \tan u = \frac{\sin ucos u}
2\int \tan u du = 2\int \frac{\sin u}{\cos u} du = 2\int \sec u du
Using the integral of secant, we have:
2\int \sec u du = 2\ln|\sec u + \tan u| + C
Substituting back u = \frac {t}{2}
2\ln|\sec\frac {t}{2} + \tan\frac {t}{2}| + C
Now, we can use the identity \sec u = \frac{1}{\cos u}
2\ln|\frac{1}{\cos\frac {t}{2}} + \tan\frac {t}{2}| +Using
\frac {2}{cos^{2}\frac {t}{2}}+C
Explanation:
The integral \int \tan\frac {t}{2}dt
can be solved using the substitution method. Let u = \frac {t}{2}
, then du = \frac {1}{2}dt
, and dt = 2du
. Substituting these into the integral, we get:
\int \tan\frac {t}{2}dt = \int \tan u \cdot 2du = 2\int \tan u du
Now, we can use the identity \tan u = \frac{\sin ucos u}
and integrate:
2\int \tan u du = 2\int \frac{\sin u}{\cos u} du = 2\int \sec u du
Using the integral of secant, we have:
2\int \sec u du = 2\ln|\sec u + \tan u| + C
Substituting back u = \frac {t}{2}
, we get:
2\ln|\sec\frac {t}{2} + \tan\frac {t}{2}| + C
Now, we can use the identity \sec u = \frac{1}{\cos u}
and simplify:
2\ln|\frac{1}{\cos\frac {t}{2}} + \tan\frac {t}{2}| +Using
\tan u = \frac{\sin u}{\cos u}, we can rewrite the expression as:
2\ln|\frac{1 + \sin\frac {t}{2}}{\cos\frac {t}{2}}| + C Now, we can use the identity
\sin^2 u + \cos^2 u = 1 to simplify the expression further:
2\ln|\frac{1 + \sin\frac {t}{2}}{\cos\frac {t}{2}}| + C = 2\ln|\frac{1 + \sin\frac {t}{2}}{\cos\frac {t}{2}}| + C Finally, we can use the identity
\ln|a| = \ln|b| + \ln|c| to separate the:
2\ln|\frac{1 + \sin\frac {t}{2}}{\cos\frac {t}{2}}| + C = 2\ln|1 + \sin\frac {t}{2}| - 2\ln|\cos\frac {t}{2}| + C Now, we can use the identity
\ln|a| - \ln|b| = \ln|\frac{a}{b}| to combine the logarithms:
2\ln|1 + \sin\frac {t}{2}| - 2\ln|\cos\frac {t}{2}| + C = 2\ln|\frac{1 + \sin\frac {t}{2}} {t}{2}}| + C Finally, we can use the identity
\ln|a| + C = \ln|a| + C to simplify the expression:
2\ln|\frac{1 + \sin\frac {t}{2}}{\cos\frac {t}{2}}| + C = \ln|\frac{1 + \sin\frac {t}{2}}{\cos\frac {t}{2}}|^2 + C Now, we can use the identity
\ln|a|^2 = \ln|a^2| to simplify the expression further:
\ln|\frac{1 + \sin\frac {t}{2}}{\cos\frac {t}{2}}|^2 + C =|\frac{(1 + \sin\frac {t}{2})^2}{\cos^2\frac {t}{2}}| + C Finally, we can use the identity
\ln|a| + C = \ln|a| + C to simplify the expression:
\ln|\frac{(1 + \sin\frac {t}{2})^2}{\cos^2\frac {t}{2}}| + C = \ln|\frac{(1 + \sin\frac {t}{2})^2}{\cos^2\frac {t}{2}}| + C Now, we can use the identity
\ln|a| + C = \ln|a| + C to simplify the expression:
\ln|\frac{(1 +\t}{2})^2}{\cos^2\frac {t}{2}}| + C = \ln|\frac{(1 + \sin\frac {t}{2})^2}{\cos^2\frac {t}{2}}| +
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