Question

Prove that ${\int }_0^{2a}f\left(x\right)dx\{\begin{array}{c}=2{\int }_0^{a}f\left(x\right)dx;f(2a-x)=f\left(x\right)\\ =0;f(2a-x)=-f\left(x\right)\end{array}$ and hence evaluate ${\int }_0^{2\pi }{\cos }^{5}xdx$.

Answer 1

Consider $I={\int }_0^{2a}f\left(x\right)dx$

$={\int }_0^{a}f\left(x\right)dx+{\int }_a^{2a}f\left(x\right)dx$

Put $x=2a-u⟹dx=-du$ in second integral

When $x=a,u=a$

When $x=2a,u=0$

$I={\int }_0^{a}f\left(x\right)dx-{\int }_a^{0}f(2a-u)du$

$⟹I={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f(2a-u)du$ ....... $\left(i\right)$

$⟹I={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(u\right)du$ ; $f(2a-x)=f\left(x\right)$

$={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$=2{\int }_0^{a}f\left(x\right)dx$

From $\left(i\right)$

$⟹I={\int }_0^{a}f\left(x\right)dx-{\int }_0^{a}f\left(u\right)du$ ; $f(2a-x)=-f\left(x\right)$

$={\int }_0^{a}f\left(x\right)dx-{\int }_0^{a}f\left(x\right)dx$

$=0$

Let $f\left(x\right)={\cos }^{5}x$

Consider $f(2\pi -x)={\cos }^5(2\pi -x)=\left(\cos \right(2\pi -x){)}^5={\cos }^{5}x$

$\therefore {\int }_0^{2\pi }{\cos }^{5}xdx$

$=2{\int }_0^{\pi }{\cos }^{5}xdx$

$={\left|\frac{{\cos }^{6}x}{6}\right(\sin x\left)\right|}_0^{\pi }$

$=0$