What is the integral from-1 to 0 of 1/(1+cos(t)) ?

The integral from-1 to 0 of 1/(1+cos(t)) is tan(1/2)

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integral from-1 to 0 of 1/(1+cos(t))

\int_{- 1}^{0} \frac{1}{1 + cos ( t )} d t

tan (\frac{1}{2})

Decimal

0.54630 \dots

Solution steps

\int_{- 1}^{0} \frac{1}{1 + cos ( t )} d t

Apply u-substitution

: \int_{- t a n (\frac{1}{2})}^{0} 1 d u

= \int_{- t a n (\frac{1}{2})}^{0} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{- t a n (\frac{1}{2})}^{0}

Simplify

= [u]_{- t a n (\frac{1}{2})}^{0}

Compute the boundaries:

tan (\frac{1}{2})

= tan (\frac{1}{2})

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What is the integral from-1 to 0 of 1/(1+cos(t)) ?
The integral from-1 to 0 of 1/(1+cos(t)) is tan(1/2)