What is the integral of sqrt(1+cos(2x-2)) ?

The integral of sqrt(1+cos(2x-2)) is sqrt(2)tan(-1+x)cos(-1+x)+C

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integral of sqrt(1+cos(2x-2))

\int 1 + cos (2 x - 2) d x

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

Solution steps

\int 1 + cos (2 x - 2) d x

Apply u-substitution

: \int 1 + cos (u) \frac{1}{2} d u

= \int 1 + cos (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 + cos (u) d u

Apply u-substitution

: \int \frac{2 2}{( 1 + v ^{2} ) 1 + v ^{2}} d v

= \frac{1}{2} \cdot \int \frac{2 2}{( 1 + v ^{2} ) 1 + v ^{2}} d v

Take the constant out:

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

= \frac{1}{2} \cdot 22 \cdot \int \frac{1}{( 1 + v ^{2} ) 1 + v ^{2}} d v

Simplify

(1 + v^{2}) 1 + v^{2} : (1 + v^{2})^{\frac{3}{2}}

= \frac{1}{2} \cdot 22 \cdot \int \frac{1}{( 1 + v ^{2} ) ^{\frac{3}{2}}} d v

Apply Trigonometric Substitution

: \int cos (w) d w

= \frac{1}{2} \cdot 22 \cdot \int cos (w) d w

Use the common integral:

\int cos (w) d w = sin (w)

= \frac{1}{2} \cdot 22 sin (w)

Substitute back

= \frac{1}{2} \cdot 22 sin (arctan (tan (\frac{2 x - 2}{2})))

Simplify

\frac{1}{2} \cdot 22 sin (arctan (tan (\frac{2 x - 2}{2}))) : 2 tan (- 1 + x) cos (- 1 + x)

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

Add a constant to the solution

= 2 tan (- 1 + x) cos (- 1 + x) + C

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What is the integral of sqrt(1+cos(2x-2)) ?
The integral of sqrt(1+cos(2x-2)) is sqrt(2)tan(-1+x)cos(-1+x)+C