What is the integral from 0 to pi/4 of sin(2x+pi/2) ?

The integral from 0 to pi/4 of sin(2x+pi/2) is 1/2

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integral from 0 to pi/4 of sin(2x+pi/2)

\int_{0}^{\frac{π}{4}} sin (2 x + \frac{π}{2}) d x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{\frac{π}{4}} sin (2 x + \frac{π}{2}) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} cos (2 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{2} d u

= \int_{0}^{\frac{π}{2}} 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_{0}^{\frac{π}{2}} cos (u) d u

Use the common integral:

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

= \frac{1}{2} [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{1}{2} \cdot 1

Simplify

= \frac{1}{2}

\int_{0}^{\frac{π}{2}} x^{2} \cdot cos (x) d x

\int_{0}^{1} x^{2} \cdot 1 - x^{2} d x

\int_{0}^{5} (4 - x^{2}) d x

\int_{e}^{6} \frac{1}{x ln ( x )} d x

\int_{1}^{2} 2 x - x^{2} d x

What is the integral from 0 to pi/4 of sin(2x+pi/2) ?
The integral from 0 to pi/4 of sin(2x+pi/2) is 1/2