Question

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

Answer 1

Hint: 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.