What is the integral from 0 to pi/2 of (x+1)sin(x) ?

The integral from 0 to pi/2 of (x+1)sin(x) is 2

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integral from 0 to pi/2 of (x+1)sin(x)

\int_{0}^{\frac{π}{2}} (x + 1) sin (x) d x

2

Solution steps

\int_{0}^{\frac{π}{2}} (x + 1) sin (x) d x

Apply Integration By Parts

: [- cos (x) (x + 1) - \int - cos (x) d x]_{0}^{\frac{π}{2}}

= [- cos (x) (x + 1) - \int - cos (x) d x]_{0}^{\frac{π}{2}}

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

= [- cos (x) (x + 1) - (- sin (x))]_{0}^{\frac{π}{2}}

Simplify

= [- cos (x) (x + 1) + sin (x)]_{0}^{\frac{π}{2}}

Compute the boundaries:

2

= 2

\int_{0}^{1} 7 x^{5} x^{3} + 8 d x

\int_{0}^{1} 24 x y d y

\int_{1}^{\infty} 4 x^{- 5} d x

\int_{0}^{1} x (x - 1) d x

\int_{0}^{2 π} sin (x + 1) d x

What is the integral from 0 to pi/2 of (x+1)sin(x) ?
The integral from 0 to pi/2 of (x+1)sin(x) is 2