Answer
The area of the region bounded by the two curves $y=tanx$ and $y=tan^{2}x$ between 0 and $\frac{\pi}{4}$ equals
$$A=\frac{1}{2}\ln{2}+\frac{\pi}{4}-1$$
Work Step by Step
The area of the region bounded by the two curves $y=tanx$ and $y=tan^{2}x$ between 0 and $\frac{\pi}{4}$ is
$$A=\int_0^\frac{\pi}{4} ({tanx -tan^{2}x})dx$$
$$A=\int_0^\frac{\pi}{4} {tanx}dx -\int_0^\frac{\pi}{4}{tan^{2}x}dx$$
The first integral is
$$ I_{1}=\int_0^\frac{\pi}{4} {tanx}dx=\int_0^\frac{\pi}{4} \frac{sinx}{cosx}dx$$
$$ I_{1}=[-\ln{cosx}]^{\frac{\pi}{4}}_{0}=$$
the second integral
$$ I_{2}=\int_0^\frac{\pi}{4}{tan^{2}x}dx$$
can be solved by rewriting the trig identity
$$ I_{2}=\int_0^\frac{\pi}{4}({-1+sec^{2}x})dx=$$
$$ I_{2}=[-x+tanx]^{\frac{\pi}{4}}_{0}$$
thus the area equals
$$A= I_{1}- I_{2}=[-\ln{cosx}]^{\frac{\pi}{4}}_{0}-[-x+tanx]^{\frac{\pi}{4}}_{0}=\frac{1}{2}\ln{2}+\frac{\pi}{4}-1$$
