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

The integral from 0 to pi/2 of sin(x)cos(y) is sin(x)

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integral from 0 to pi/2 of sin(x)cos(y)

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

sin (x)

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= sin (x) [sin (y)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= sin (x) \cdot 1

Simplify

= sin (x)

\int_{2}^{x^{2}} t^{2} t^{2} + 9 d t

\int_{0}^{10 π} ∣ sin (x) ∣ d x

\int_{0}^{2} sin (\frac{nπ}{2} x) d x

\int_{0}^{\frac{5}{3}} 3 x^{2} - 5 x d x

\int_{3}^{x} e^{t^{2}} d t

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