#### To determine

**To show:**

∫0π2f(cosx)dx=∫0π2f(sinx)dx

#### Answer

∫0π2f(cosx)dx=∫0π2f(sinx)dx

#### Explanation

**1) Concept:**

i. The substitution rule: If u=g(x) is a differentiable function whose range is I, and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du. Here g(x) is substituted as u and then differentiation g’(x)dx =du

ii.

∫abf(x)dx=-∫baf(x)dx

iii.

∫abf(u)du=∫abf(x)dx

**2) Given:**

f is continuous

**3) Formula:**

cosπ2-x=sinx

**4) calculation:**

Let

∫0π2f(cosx)dx

Here use the substitution method

Substitute π2-x=u

That is the same as x=π2-u

Differentiating with respect to x

dx=-du

The limits changes; the new limits of integration are calculated by substituting

At x=0, u=π2-0=π2 and

At x=π2, u=π2-π2=0

Therefore, the integral becomes

∫0π2fcosxdx

=∫π20fcosπ2-u(-du)

Simplify and using cosπ2-x=sinx

=∫π20-f(sinu)du

Usingthe property of definite integral

=∫0π2f(sinu)du

Using concept

=∫0π2f(sinx)dx

Therefore,

∫0π2f(cosx)dx=∫0π2f(sinx)dx

**Conclusion:**

Therefore,

∫0π2f(cosx)dx=∫0π2f(sinx)dx