Solveeit Logo
Login

Question

Question: Let \( {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx,} \) then \( \dfrac{1}{{{I_2} + {I_4}}},\dfr...

Let In=∫0π4tan⁡nxdx,{I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx,} then 1I2+I4,1I3+I5,1I4+I6,1I5+I7,\dfrac{1}{{{I_2} + {I_4}}},\dfrac{1}{{{I_3} + {I_5}}},\dfrac{1}{{{I_4} + {I_6}}},\dfrac{1}{{{I_5} + {I_7}}}, form an:
A) AP with common difference 22
B) GP with common ratio 22
C) AP with common difference 11
D) HP

Explanation

Solution

Hint : To solve this question, we will integrate the expression given to us, i.e., In=∫0π4tan⁡nxdx,{I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx,} then after solving it, we will get an equation in the form, 1In+In−2=(n−1).\dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1). Now, on substituting the values of n = 4,5,6,7…in this equation, we will check whether we will get AP, GP or HP.

Complete step-by-step answer :
n a sequence of numbers when the difference between any two consecutive terms is same, then the sequence is an arithmetic progression, while if the sequence of numbers have the ratio of any two consecutive terms is same, then the sequence is a geometric progression, if in the sequence of numbers, the reciprocal of the terms are in AP, then it is a harmonic progression.
We have been given that, In=∫0π4tan⁡nxdx,{I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx,} we need to check the expression 1I2+I4,1I3+I5,1I4+I6,1I5+I7,\dfrac{1}{{{I_2} + {I_4}}},\dfrac{1}{{{I_3} + {I_5}}},\dfrac{1}{{{I_4} + {I_6}}},\dfrac{1}{{{I_5} + {I_7}}}, will form an AP, GP or HP.
In=∫0π4tan⁡nxdx{I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx}
In=∫0π4tan⁡n−2x.tan⁡2xdx{I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x.{{\tan }^2}xdx}
In=∫0π4tan⁡n−2x(sec⁡2x−1)dx{I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x({{\sec }^2}x - 1)dx} [∵tan⁡2x=sec⁡2x−1][\because {\tan ^2}x = {\sec ^2}x - 1]
In=∫0π4(tan⁡n−2x.sec⁡2x)dx−∫0π4tan⁡n−2xdx{I_n} = \int_0^{\dfrac{\pi }{4}} {({{\tan }^{n - 2}}x.{{\sec }^2}x)dx - \int_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}xdx} }
In=∫0π4(tan⁡n−2x.sec⁡2x)dx−In−2{I_n} = \int_0^{\dfrac{\pi }{4}} {({{\tan }^{n - 2}}x.{{\sec }^2}x)dx - } {I_{n - 2}} [∵In=∫0π4tan⁡nxdx,∴In−2=∫0π4tan⁡n−2xdx][\because {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx} ,\therefore {I_{n - 2}} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}xdx} ]
In+In−2=∫0π4(tan⁡n−2x.sec⁡2x)dx{I_n} + {I_{n - 2}} = \int_0^{\dfrac{\pi }{4}} {({{\tan }^{n - 2}}x.{{\sec }^2}x)dx}
In+In−2=(tn−1n−1)01{I_n} + {I_{n - 2}} = (\dfrac{{{t^{n - 1}}}}{{n - 1}})_0^ 1
In+In−2=(1n−1){I_n} + {I_{n - 2}} = (\dfrac{1}{{n - 1}})
⇒1In+In−2=(n−1)\Rightarrow \dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1)
Now, let us check this expression, 1I2+I4,1I3+I5,1I4+I6,1I5+I7.\dfrac{1}{{{I_2} + {I_4}}},\dfrac{1}{{{I_3} + {I_5}}},\dfrac{1}{{{I_4} + {I_6}}},\dfrac{1}{{{I_5} + {I_7}}}.
On putting n=4 in 1In+In−2=(n−1),\dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), we get
1I4+I4−2=1I2+I4=(4−1)=3\dfrac{1}{{{I_4} + {I_{4 - 2}}}} = \dfrac{1}{{{I_2} + {I_4}}} = (4 - 1) = 3
On putting n=5 in 1In+In−2=(n−1),\dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), we get
1I5+I5−2=1I3+I5=(5−1)=4\dfrac{1}{{{I_5} + {I_{5 - 2}}}} = \dfrac{1}{{{I_3} + {I_5}}} = (5 - 1) = 4
On putting n=6 in 1In+In−2=(n−1),\dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), we get
1I6+I6−4=1I4+I6=(6−1)=5\dfrac{1}{{{I_6} + {I_{6 - 4}}}} = \dfrac{1}{{{I_4} + {I_6}}} = (6 - 1) = 5
On putting n=7 in 1In+In−2=(n−1),\dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), we get
1I7+I7−2=1I5+I7=(7−1)=6\dfrac{1}{{{I_7} + {I_{7 - 2}}}} = \dfrac{1}{{{I_5} + {I_7}}} = (7 - 1) = 6
So, we get 3,4,5,6, we can see the common difference between two consecutive terms is 1. Hence, 3,4,5,6 are in AP, with a common difference 1.
Thus, option (C) AP with common difference 11.
So, the correct answer is “Option C”.

Note : Students should note that in these types of questions, always calculate the expression carefully, because if you make the mistake in evaluating that, you will end up getting your answer wrong. So, here, getting this 1In+In−2=(n−1),\dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), term correct was necessary.