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TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Section: Mathematics

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Question : 75 of 160

Marks: +1, -0

Let

\tan^0 x

1

\int\left(\sum\limits_{k=0}^{7}\tan^k x\right)dx

\sum\limits_{k=1}^{7} A_k \tan^k x + C

\sum\limits_{k=1}^{7}

A_k =

$\frac{76}{25}$
$\frac{28}{15}$
$\frac{38}{35}$
$\frac{124}{75}$

Solution:

Given,

\tan^{0} x = 1

\int(\tan^{0}x + \tan x + \tan^{2}x + \tan^{3}x + \dots + \tan^{7}x)dx =

A_{1}\tan x + A_{2}\tan^{2}x + \dots + A_{7}\tan^{7}x

. . . (i)

If

I_{n} = \int\tan^{n}x\,dx \Rightarrow I_{n}+I_{n-2} = \frac{(\tan x)^{n-1}}{n-1}

Put

n=2,\; I_{0}+I_{2}=\tan x \Rightarrow n=3,\; I_{1}+I_{3}=\frac{\tan^{2}x}{2}

n=6,\; I_{4}+I_{6}=\frac{\tan^{5}x}{5} \Rightarrow n=7,\; I_{5}+I_{7}=\frac{\tan^{6}x}{6}

Put

n=2,\; I_{0}+I_{2}=\tan x \Rightarrow n=3,\; I_{1}+I_{3}=\frac{\tan^{2}x}{2}

From Eq. (i), we get

I_{0}+I_{1}+I_{2}+I_{3}+\dots+I_{7}

= \tan x + \frac{\tan^{2}x}{2} + \frac{\tan^{5}x}{5} + \dots + \frac{\tan^{6}x}{6}

On comparing, we get

A_{1}=1,\; A_{2}=\frac{1}{2},\; A_{3}=A_{4}=A_{7}=0,\; A_{5}=\frac{1}{5}

A_{6}=\frac{1}{6}

\sum\limits_{k=1}^{7} A_{k} = A_{1}+A_{2}+A_{3}+\dots+A_{7}

= 1 + \frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{56}{30} = \frac{28}{15}

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