Answer
$$\int^{\pi/4}_0\frac{\sec^2x}{(1+7\tan x)^{2/3}}dx=\frac{3}{7}$$
Work Step by Step
$$A=\int^{\pi/4}_0\frac{\sec^2x}{(1+7\tan x)^{2/3}}dx$$ $$A=\int^{\pi/4}_0\sec^2x(1+7\tan x)^{-2/3}dx$$
We set $u=1+7\tan x$, which means $$du=7\sec^2xdx$$ $$\sec^2xdx=\frac{1}{7}du$$
For $x=\pi/4$: $$u=1+7\tan(\pi/4)=1+7\times1=8$$
For $x=0$: $$u=1+7\tan0=1+7\times0=1$$
Therefore, $$A=\frac{1}{7}\int^8_1u^{-2/3}du$$ $$A=\Big(\frac{1}{7}\times\frac{u^{1/3}}{\frac{1}{3}}\Big)\Big]^8_1$$ $$A=\Big(\frac{3\sqrt[3]u}{7}\Big)\Big]^8_1$$ $$A=\frac{3}{7}(2-1)=\frac{3}{7}$$
