Question

For each pair of definite integrals in exercise 13-18 decide which if either is larger without computing the integrals

$\pi {\int }_0^{\frac{\pi }{2}}{\cos }^2\left(x\right)dxand\pi {\int }_{\frac{\pi }{2}}^{\pi }{\cos }^2\left(x\right)dx$

Short answer

Both integrals are equal

Step-by-step solution

Given information

We are given two integrals$\pi {\int }_0^{\frac{\pi }{2}}{\cos }^2\left(x\right)dxand\pi {\int }_{\frac{\pi }{2}}^{\pi }{\cos }^2\left(x\right)dx$

Explaination

We will substitute $\cos \left(x^2\right)=\frac{1-\cos 2x}{2}$

Hence the integral becomes

$\pi {\int }_0^{\frac{\pi }{2}}\frac{1-\cos 2x}{2}dxand\pi {\int }_{\frac{\pi }{2}}^{\pi }\frac{1-\cos 2x}{2}dx$

On integrating we will get sine terms

And after substituting the limits we will get equal values