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Sonja [21]
2 years ago
12

Evaluate this equation

Mathematics
1 answer:
alexgriva [62]2 years ago
4 0

Given the integral

\displaystyle \int \tan(3x) \, dx

first substitute y = 3x and dy = 3 dx :

\displaystyle \int \tan(3x) \, dx = \frac13 \int \tan(y) \, dy

then rewrite tan(y) = sin(y)/cos(y) and substitute z = cos(y) and dz = -sin(y) dy :

\displaystyle \frac13 \int \frac{\sin(y)}{\cos(y)} \, dy = -\frac13 \int \frac{dz}z

Recall that 1/z is the derivative of ln|z|. Then back-substitute to get the result in terms of x :

\displaystyle -\frac13 \int \frac{dz}z = -\frac13 \ln|z| + C

\displaystyle \frac13 \int \tan(y) \, dy = -\frac13 \ln|\cos(y)| + C

\displaystyle \boxed{\int \tan(3x) \, dx = -\frac13 \ln|\cos(3x)| + C}

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