What is the integral from 0 to 6 of 9sin(sqrt(x+1)) ?

The integral from 0 to 6 of 9sin(sqrt(x+1)) is 18(-sqrt(7)cos(sqrt(7))+sin(sqrt(7))+cos(1)-sin(1))

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integral from 0 to 6 of 9sin(sqrt(x+1))

\int_{0}^{6} 9 sin (x + 1) d x

18 (- 7 cos (7) + sin (7) + cos (1) - sin (1))

Decimal Notation

45.03101 \dots

Solution steps

\int_{0}^{6} 9 sin (x + 1) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{6} sin (x + 1) d x

Apply u-substitution

: \int_{1}^{7} sin (u) \cdot 2 u d u

= 9 \cdot \int_{1}^{7} sin (u) \cdot 2 u d u

Take the constant out:

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

= 9 \cdot 2 \cdot \int_{1}^{7} sin (u) u d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{1}^{7}

= 9 \cdot 2 [- u cos (u) - \int - cos (u) d u]_{1}^{7}

\int - cos (u) d u = - sin (u)

= 9 \cdot 2 [- u cos (u) - (- sin (u))]_{1}^{7}

Simplify

= 18 [- u cos (u) + sin (u)]_{1}^{7}

Compute the boundaries:

- 7 cos (7) + sin (7) + cos (1) - sin (1)

= 18 (- 7 cos (7) + sin (7) + cos (1) - sin (1))

- 4 (8 x^{2} + 8)^{- 6}

(1 + x)^{^{'}}

f (x) = \frac{1}{7 ( 1 + x ^{2} )}, g (x) = \frac{1}{14} x^{2}

\int \frac{1 + 13 x}{x} d x

x \cdot y \cdot y^{^{'}} = y^{2} + 1

What is the integral from 0 to 6 of 9sin(sqrt(x+1)) ?
The integral from 0 to 6 of 9sin(sqrt(x+1)) is 18(-sqrt(7)cos(sqrt(7))+sin(sqrt(7))+cos(1)-sin(1))