If $\int_0^{\pi / 2}\left(\sin ^4 x+\cos ^4 x\right) d x=k$, then what is the value of $\int_0^{20 \pi}\left(\sin ^4+\cos ^4 x\right) d x ?$

Concept: If $\mathrm{f}(\mathrm{x})$ is a periodic function with period T , then $\int_0^{n T} f(x) d x=n \int_0^T f(x) d x$ If $f(x)$ is periodic with period $T$, then $f(a x)$ is periodic with period $\frac{T}{|a|}$ Calculation: $\begin{aligned} & \text { Let } f(x)=\left(\sin ^4 x+\cos ^4 x\right) \\ & \Rightarrow f(x)=\left(\sin ^2 x\right)^2+\left(\cos ^2 x\right)^2+2 \sin ^2 x \cos ^2 x-2 \sin ^2 x \cos ^2 x \\ & \Rightarrow f(x)=\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x \\ & \Rightarrow f(x)=1-\frac{\sin ^2 2 x}{2} \\ & \Rightarrow f(x)=1-\frac{1-\cos 4 x}{4} \\ & \Rightarrow f(x)=\frac{3}{4}+\frac{\cos 4 x}{4} \end{aligned}$ Since $\cos x$ is a function of period $2 \pi$ $\Rightarrow \cos 4 x$ is periodic with period $\frac{2 \pi}{4}=\frac{\pi}{2}$ $\Rightarrow f(x)=\left(\sin ^4 x+\cos ^4 x\right)$ is periodic with period $\frac{\pi}{2}$ Given, $\int_0^{\pi / 2}\left(\sin ^4 x+\cos ^4 x\right) d x=k$ $\qquad$ So, $\int_0^{20 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x=\int_0^{40\left(\frac{\pi}{2}\right)}\left(\sin ^4 x+\cos ^4 x\right) d x$ As $f(x)=\left(\sin ^4 x+\cos ^4 x\right)$ is periodic with period $\frac{\pi}{2}$ By the property of definite integral, $\begin{aligned} & \int_0^{40\left(\frac{\pi}{2}\right)}\left(\sin ^4 x+\cos ^4 x\right) d x=40 \int_0^{\pi / 2}\left(\sin ^4 x+\cos ^4 x\right) d x \\ & \Rightarrow \int_0^{20 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x=40 k\{\text { Using }(i)\} \end{aligned}$