What is maclaurin ln(1+cos(x)) ?

The answer to maclaurin ln(1+cos(x)) is ln(2)-1/4 x^2-1/96 x^4-1/1440 x^6+\ldots

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maclaurin ln(1+cos(x))

maclaurin

ln (1 + cos (x))

ln (2) - \frac{1}{4} x^{2} - \frac{1}{96} x^{4} - \frac{1}{1440} x^{6} + \dots

Solution steps

Apply the Maclaurin Formula

= ln (2) + \frac{\frac{d}{d x} ( ln ( 1 + cos ( x ) ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 + cos ( x ) ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 + cos ( x ) ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= ln (2) + \frac{0}{1 !} x + \frac{- \frac{1}{2}}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- \frac{1}{4}}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- \frac{1}{2}}{6 !} x^{6} + \dots

Refine

= ln (2) - \frac{1}{4} x^{2} - \frac{1}{96} x^{4} - \frac{1}{1440} x^{6} + \dots

2 y y^{^{'}} = x (y^{2} + 1), y (2) = 0

f (x) = x^{4} cos (x)

\int e^{- 2 x} d x

\frac{d y}{d t} + y = t

\int 18 cos (x) d x

What is maclaurin ln(1+cos(x)) ?
The answer to maclaurin ln(1+cos(x)) is ln(2)-1/4 x^2-1/96 x^4-1/1440 x^6+\ldots