\(\int_0^{\pi / 4}\left(\tan ^2 x-\tan ^4 x\right) d x=\)
\(\int_0^{\pi / 4}\left(\tan ^2 x-\tan ^4 x\right) d x=\)
3
2
\(\frac{1}{3}\)
0
Solution
\(\begin{aligned}
& \int_0^{\frac{\pi}{4}}\left(\tan ^2 x+\tan ^4 x\right) d x \\
& \int_0^{\frac{\pi}{4}}\left(1+\tan ^2 x\right) \tan ^2 x d x \\
& \int_0^{\frac{\pi}{4}} \sec ^2 x \cdot \tan ^2 x d x \\
& \text {Put, } \tan x=t \\
& \sec ^2 x d x=d t
\end{aligned}\)
when \(x=\frac{\pi}{4}\), then \(t=1\) and when \(x=0\), then \(t=0\)
\(=\int_0^1 t^2 \cdot d t=\left(\frac{t^3}{3}\right)_0^1=\left(\frac{1}{3}-0\right)=\frac{1}{3}\)
Hence, option (c) is correct.