Question

Find the integrals of the functions.
$\int ta{n}^{4}xdx.$

Answer 1

$\int ta{n}^{4}xdx$
Let $I=\int ta{n}^{4}xdx=\int (ta{n}^{2}x{)}^{2}dx\Rightarrow I=\int (ta{n}^{2}x\left)\right(ta{n}^{2}x)dx$
$=\int (se{c}^{2}x-1)\left(ta{n}^{2}x\right)dx=\int se{c}^{2}xta{n}^{2}xdx-\int ta{n}^{2}xdx=\int se{c}^{2}xta{n}^{2}xdx-\int [se{c}^{2}x-1]dx=\int se{c}^{2}xta{n}^{2}xdx-[\int se{c}^{2}xdx-\int 1dx]$
Now, let $I_1=\int se{c}^{2}xta{n}^{2}xdxand{I}_2=\int se{c}^{2}xdx-\int 1dx$
Then, $I=I_1-I_2.......\left(i\right)$
Putting tan $x=t\Rightarrow se{c}^{2}x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{se{c}^{2}x}$

$\therefore I_1=\int se{c}^{2}x{t}^2.\frac{dt}{se{c}^{2}x}\Rightarrow I_1=\int t^{2}dt=\frac{t^3}{3}+C_1=\frac{ta{n}^{3}x}{3}+C_1{I}_2=\int se{c}^{2}xdx-\int 1dx=tanx-x-C_2\therefore \text{Putting the values of I\_1 and I\_2 in Eq. (i), we get}I=\frac{ta{n}^{3}x}{3}+C_1-(tanx-x)+C_2\Rightarrow I=\frac{ta{n}^{3}x}{3}-tanx+x+C(\because constant+constant=constant\therefore C_1+C_2=C)$