What is the value of \(\rm \displaystyle\int_0^{\pi/4}(\tan^3 x + \tan x)dx \ ?\)

Concept:

1 + tan² x = sec² x

\(\rm \displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+c\)

Calculation:

\(\text{Let I} = \rm \displaystyle \int_0^{π/4} (\tan^3 x + \tan x)dx\\\Rightarrow\rm \displaystyle \int_0^{π/4} \tan x(\tan^2 x + 1)dx\)

\(\\\Rightarrow \rm \displaystyle \int_0^{π/4} \tan x \sec^2 x dx\)             (∵ 1 + tan² x = sec² x)

Let tanx = t

Differentiating with respect to x, we get

⇒ sec2 x dx = dt

\(\Rightarrow\rm \displaystyle \int_0^1 tdt\\\Rightarrow[\frac{t^2}{2}]_0^1\\\Rightarrow\frac{1}{2}-0=\frac{1}{2}\)

∴ The value of the integral \(\rm \displaystyle\int_0^{\pi/4}(\tan^3 x + \tan x)dx \ \)is 1/2.