Question

If \[{u_n} = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx} \], then what is the value of \[{u_n} + {u_{n - 2}}\]?
A. \[\dfrac{1}{{n - 1}}\]
B. \[\dfrac{1}{{n + 1}}\]
C. \[\dfrac{1}{{2n - 1}}\]
D. \[\dfrac{1}{{2n + 1}}\]

Answer 1

Hint: Here, a definite integral is given. First, rewrite the function \[{\tan ^n}x = {\tan ^{n - 2}}x{\tan ^2}x\]. Then, simplify the function by applying the trigonometric formula \[{\tan ^2}x = {\sec ^2}x - 1\]. After that, separate the terms by applying the integration formula \[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\]. In the end, solve the integral by substitution method and apply the limits to get the required answer.



Formula Used:\[{\tan ^2}x = {\sec ^2}x - 1\]
\[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\]
Integration rule: \[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]



Complete step by step solution:The given definite integral is \[{u_n} = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx} \].
Rewrite \[{\tan ^n}x = {\tan ^{n - 2}}x{\tan ^2}x\] in the above integral.
\[{u_n} = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x{{\tan }^2}xdx} \]
Simplify the function by using the trigonometric formula \[{\tan ^2}x = {\sec ^2}x - 1\].
\[{u_n} = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)dx} \]
\[ \Rightarrow {u_n} = \int\limits_0^{\dfrac{\pi }{4}} {\left[ {{{\tan }^{n - 2}}x{{\sec }^2}x - {{\tan }^{n - 2}}x} \right]dx} \]
Separate the terms by applying the integration rule \[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\].
\[ \Rightarrow {u_n} = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x{{\sec }^2}xdx} - \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}xdx} \]
From the given equation we get \[{u_{n - 2}} = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}xdx} \].
So,
\[ \Rightarrow {u_n} = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x{{\sec }^2}xdx} - {u_{n - 2}}\]
\[ \Rightarrow {u_n} + {u_{n - 2}} = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x{{\sec }^2}xdx} \] \[.....\left( 1 \right)\]
Now substitute \[v = \tan x\] in the above integral.
Differentiate the substituted integral.
We get,
\[dv = {\sec ^2}xdx\]
Now calculate the changed limits of the integral.
At \[x = 0\]:
\[v = \tan 0\]
\[ \Rightarrow v = 0\]

At \[x = \dfrac{\pi }{4}\]:
\[v = \tan \dfrac{\pi }{4}\]
\[ \Rightarrow v = 1\]

So, we get the equation \[\left( 1 \right)\] as follows:
\[ \Rightarrow {u_n} + {u_{n - 2}} = \int\limits_0^1 {{v^{n - 2}}dv} \]
Solve the integral by using the integration formula \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
\[ \Rightarrow {u_n} + {u_{n - 2}} = \left[ {\dfrac{{{v^{n - 2 + 1}}}}{{n - 2 + 1}}} \right]_0^1\]
\[ \Rightarrow {u_n} + {u_{n - 2}} = \left[ {\dfrac{{{v^{n - 1}}}}{{n - 1}}} \right]_0^1\]
Apply the limits.
\[ \Rightarrow {u_n} + {u_{n - 2}} = \dfrac{{{1^{n - 1}}}}{{n - 1}} - \dfrac{{{0^{n - 1}}}}{{n - 1}}\]
\[ \Rightarrow {u_n} + {u_{n - 2}} = \dfrac{1}{{n - 1}} - \dfrac{0}{{n - 1}}\]
\[ \Rightarrow {u_n} + {u_{n - 2}} = \dfrac{1}{{n - 1}} - 0\]
\[ \Rightarrow {u_n} + {u_{n - 2}} = \dfrac{1}{{n - 1}}\]



Option ‘A’ is correct



Note: Students often do mistake and calculate the separate value of \[{u_{n - 2}}\] by using the given integral. Then, they add both integrals and try to solve them. Because of that, they get the wrong solution.