What is the integral from 1 to x of sqrt(cos(t)-1) ?

The integral from 1 to x of sqrt(cos(t)-1) is i(2sqrt(1+cos(1))-2sqrt(1+cos(x)))

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integral from 1 to x of sqrt(cos(t)-1)

\int_{1}^{x} cos (t) - 1 d t

i (2 1 + cos (1) - 2 1 + cos (x))

Solution steps

\int_{1}^{x} cos (t) - 1 d t

Apply u-substitution

: \int_{c o s (1)}^{c o s (x)} - i \frac{1}{u + 1} d u

= \int_{c o s (1)}^{c o s (x)} - i \frac{1}{u + 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - i \cdot \int_{c o s (1)}^{c o s (x)} \frac{1}{u + 1} d u

Apply u-substitution

: \int_{c o s (1) + 1}^{c o s (x) + 1} \frac{1}{v} d v

= - i \cdot \int_{c o s (1) + 1}^{c o s (x) + 1} \frac{1}{v} d v

Simplify

\frac{1}{v} : \frac{1}{v}

= - i \cdot \int_{c o s (1) + 1}^{c o s (x) + 1} \frac{1}{v} d v

Simplify

= - i \cdot \int_{c o s (1) + 1}^{c o s (x) + 1} \frac{1}{v} d v

Apply the Power Rule

: [2 v^{\frac{1}{2}}]_{c o s (1) + 1}^{c o s (x) + 1}

= - i [2 v^{\frac{1}{2}}]_{c o s (1) + 1}^{c o s (x) + 1}

Apply radical rule:

= - i [2 v]_{c o s (1) + 1}^{c o s (x) + 1}

Compute the boundaries:

2 cos (x) + 1 - 2 cos (1) + 1

= - i (2 cos (x) + 1 - 2 cos (1) + 1)

Simplify

= i (2 1 + cos (1) - 2 1 + cos (x))

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What is the integral from 1 to x of sqrt(cos(t)-1) ?
The integral from 1 to x of sqrt(cos(t)-1) is i(2sqrt(1+cos(1))-2sqrt(1+cos(x)))