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

Question

Answer 1

The correct answer is:

Explanation:The integral

can be solved using the substitution method. Let

, then

, and

. Substituting these into the integral, we get:

Now, we can use the identity

and integrate:

Using the integral of secant, we have:

Substituting back

, we get:

Now, we can use the identity

and simplify:

\tan u = \frac{\sin u}{\cos u}

2\ln|\frac{1 + \sin\frac {t}{2}}{\cos\frac {t}{2}}| + C

\sin^2 u + \cos^2 u = 1

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

\ln|a| = \ln|b| + \ln|c|

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

\ln|a| - \ln|b| = \ln|\frac{a}{b}|

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

\ln|a| + C = \ln|a| + C

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

\ln|a|^2 = \ln|a^2|

\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

\ln|a| + C = \ln|a| + C

\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

\ln|a| + C = \ln|a| + C

\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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