What is 1/4 (ln(3tan(x/2+1))-ln(tan(x/2+3)))+c ?

The solution to 1/4 (ln(3tan(x/2+1))-ln(tan(x/2+3)))+c is 1/4 ln(3tan(x/2+1)cot(x/2+3))+c

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simplify 1/4 (ln(3tan(x/2+1))-ln(tan(x/2+3)))+c

simplify

\frac{1}{4} (ln (3 tan (\frac{x}{2} + 1)) - ln (tan (\frac{x}{2} + 3))) + c

\frac{1}{4} ln (3 tan (\frac{x}{2} + 1) cot (\frac{x}{2} + 3)) + c

Solution steps

\frac{1}{4} (ln (3 tan (\frac{x}{2} + 1)) - ln (tan (\frac{x}{2} + 3))) + c

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (3 tan (\frac{x}{2} + 1)) - ln (tan (\frac{x}{2} + 3)) = ln (\frac{3 tan ( \frac{x}{2} + 1 )}{tan ( \frac{x}{2} + 3 )})

= \frac{1}{4} ln (\frac{3 tan ( \frac{x}{2} + 1 )}{tan ( \frac{x}{2} + 3 )}) + c

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

= \frac{1}{4} ln (3 tan (\frac{x}{2} + 1) cot (\frac{x}{2} + 3)) + c

0.423 - \frac{0.0592}{2} lo g_{10} (\frac{0.075}{0.1})

150 ln (\frac{350}{300})

ln (∣ cos (θ) 2 (ln ∣ 1 + cos (2 θ) ∣ - ln (2)))

ln (e^{3 x} \cdot sin (2 x))

ln (- 2 \cdot e^{- x})

What is 1/4 (ln(3tan(x/2+1))-ln(tan(x/2+3)))+c ?
The solution to 1/4 (ln(3tan(x/2+1))-ln(tan(x/2+3)))+c is 1/4 ln(3tan(x/2+1)cot(x/2+3))+c