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

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

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

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

f

Solution steps

\int_{0}^{\frac{π}{2}} f sin (x) d x

Take the constant out:

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

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

Use the common integral:

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

= f [- cos (x)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= f \cdot 1

Simplify

= f

\int_{0}^{1} \frac{x}{1 - x} d x

\int_{e}^{e^{2}} \frac{ln ( x )}{x} d x

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{3}} d x

\int_{0}^{3} x e^{2 x} d x

\int_{0}^{2} - x^{2} + 2 x d x

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