Question

Obtain reduction formula: $\int {{{\tan }^n}xdx}$ for integer $n \geqslant 2$ and evaluate $\int {{{\tan }^6}xdx}$

Answer 1

First we will assume the given integral to be equal to I and then simplify it and use the following identity:-
${\tan ^2}x = {\sec ^2}x - 1$
And also, the formulas for integration:-
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}$
$\int {{{\sec }^2}xdx = \tan x + C}$
And for evaluating the value of $\int {{{\tan }^6}xdx}$ we will substitute the value of n as 6 in the formula obtained.

Complete step-by-step answer:
Let ${I_n} = \int {{{\tan }^n}xdx}$
Simplifying it we get:-
${I_n} = \int {{{\tan }^2}x{{\tan }^{n - 2}}xdx}$
Now we know that,
${\tan ^2}x = {\sec ^2}x - 1$
Hence applying this identity we get:-
${I_n} = \int {\left( {{{\sec }^2}x - 1} \right){{\tan }^{n - 2}}xdx}$
Simplifying it further we get:-
${I_n} = \int {{{\sec }^2}x} {\tan ^{n - 2}}xdx - \int {{{\tan }^{n - 2}}xdx}$
Now we know that,
${I_{n - 2}} = \int {{{\tan }^{n - 2}}xdx}$
Hence, substituting this value in above equation we get:-
${I_n} = \int {{{\sec }^2}x} {\tan ^{n - 2}}xdx - {I_{n - 2}}$
Simplifying it we get:-
${I_n} + {I_{n - 2}} = \int {{{\sec }^2}x} {\tan ^{n - 2}}xdx$………………………. (1)
Now let $\tan x = u$
Differentiating both the sides with respect to x we get:-
$\dfrac{d}{{dx}}\left( {\tan x} \right) = \dfrac{{du}}{{dx}}$
We know that, $\dfrac{d}{{dx}}\tan x = {\sec ^2}x$
Hence, substituting the value we get:-
${\sec ^2}x = \dfrac{{du}}{{dx}}$
$\Rightarrow {\sec ^2}xdx = du$
Substituting the respective values in equation 1 we get:-
${I_n} + {I_{n - 2}} = \int {{u^{n - 2}}du}$
Now we know that,
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}$
Hence, applying this formula we get:-
${I_n} + {I_{n - 2}} = \dfrac{{{u^{n - 2 + 1}}}}{{n - 2 + 1}} + C$
Simplifying it we get:-
${I_n} + {I_{n - 2}} = \dfrac{{{u^{n - 1}}}}{{n - 1}} + C$
Now substituting back the value of u we get:-
${I_n} + {I_{n - 2}} = \dfrac{{{{\tan }^{n - 1}}x}}{{n - 1}} + C$
Therefore, the reduction formula is:-
${I_n} = \dfrac{{{{\tan }^{n - 1}}x}}{{n - 1}} - {I_{n - 2}}$
Now we need to evaluate the value of $\int {{{\tan }^6}xdx}$
Hence, we need to substitute the value of n as 6.
Therefore, on substituting we get:-
${I_6} = \dfrac{{{{\tan }^{6 - 1}}x}}{{6 - 1}} - {I_{6 - 2}}$
Simplifying it we get:-
${I_6} = \dfrac{{{{\tan }^5}x}}{5} - {I_4}$……………………….. (2)
Now we need to find the value of ${I_4}$
Therefore, putting $n = 4$in reduction formula we get:-
${I_4} = \dfrac{{{{\tan }^{4 - 1}}x}}{{4 - 1}} - {I_{4 - 2}}$
Simplifying it we get:-
${I_4} = \dfrac{{{{\tan }^3}x}}{3} - {I_2}$……………………. (3)
Now we have to find the value of ${I_2}$
Therefore, putting $n = 2$in reduction formula we get:-
${I_2} = \dfrac{{{{\tan }^{2 - 1}}x}}{{2 - 1}} - {I_{2 - 2}}$
Simplifying it we get:-
${I_2} = \tan x - {I_0}$……………………. (4)
Now we know that,
${I_0} = \int {{{\tan }^0}x} dx$
${I_0} = \int 1 dx$
On integration we get:-
${I_0} = x$
Substituting this value in equation 4 we get:-
${I_2} = \tan x - x$
Now substituting this value in equation 3 we get:-
${I_4} = \dfrac{{{{\tan }^3}x}}{3} - \left( {\tan x - x} \right)$
$\Rightarrow {I_4} = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x$
Now substituting this value in equation 2 we get:-
${I_6} = \dfrac{{{{\tan }^5}x}}{5} - \left( {\dfrac{{{{\tan }^3}x}}{3} - \tan x + x} \right)$
Simplifying it we get:-

${I_6} = \dfrac{{{{\tan }^5}x}}{5} - \dfrac{{{{\tan }^3}x}}{3} + \tan x - x + C$.

Note: Students might forget to substitute back the value of u while finding the reduction formula.
Also, in finding values like ${I_6}$ we should use the reduction formula to ease the calculations.
Students should note that, the general formula for integration is given by:-
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}$
Also, never forget to add a constant of integration C after integration in case of indefinite integrals.