What is the integral of sqrt(162+162cos(x)) ?

The integral of sqrt(162+162cos(x)) is 36tan(x/2)cos(x/2)+C

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integral of sqrt(162+162cos(x))

\int 162 + 162 cos (x) d x

36 tan (\frac{x}{2}) cos (\frac{x}{2}) + C

Solution steps

\int 162 + 162 cos (x) d x

162 + 162 cos (x) = 162 1 + cos (x)

= \int 162 1 + cos (x) d x

Take the constant out:

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

= 162 \cdot \int 1 + cos (x) d x

162 = 92

= 92 \cdot \int 1 + cos (x) d x

Apply u-substitution

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

= 92 \cdot \int \frac{2 2}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Take the constant out:

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

= 92 \cdot 22 \cdot \int \frac{1}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Simplify

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

= 92 \cdot 22 \cdot \int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int cos (v) d v

= 92 \cdot 22 \cdot \int cos (v) d v

Use the common integral:

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

= 92 \cdot 22 sin (v)

Substitute back

= 92 \cdot 22 sin (arctan (tan (\frac{x}{2})))

Simplify

92 \cdot 22 sin (arctan (tan (\frac{x}{2}))) : 36 tan (\frac{x}{2}) cos (\frac{x}{2})

= 36 tan (\frac{x}{2}) cos (\frac{x}{2})

Add a constant to the solution

= 36 tan (\frac{x}{2}) cos (\frac{x}{2}) + C

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What is the integral of sqrt(162+162cos(x)) ?
The integral of sqrt(162+162cos(x)) is 36tan(x/2)cos(x/2)+C