Q115 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics
Let $\tan ^0 x=1$. If $\int\left(\sum_{k=0}^7 \tan ^k x\right) d x=\sum_{k=1}^7 A_k \tan ^k x+C$, then $\sum_{k=1}^7 A_k$ is equal to
- A76/25
- Correct answerB28/15
- C38/35
- D124/75
Solution
Given, $\tan ^0 x=1$
$$
\int\left(\tan ^0 x+\tan x+\tan ^2 x+\tan ^3 x+\ldots+\tan ^7 x\right) d x=
$$
If $I_n=\int \tan ^n x d x \Rightarrow I_n+I_{n-2}=\frac{(\tan x)^{n-1}}{n-1}$ Put $n=2, I_0+I_2=\tan x \Rightarrow n=3, I_1+I_3=\frac{\tan ^2 x}{2}$ $$ n=6, I_4+I_6=\frac{\tan ^5 x}{5} \Rightarrow n=7, I_5+I_7=\frac{\tan ^6 x}{6} $$ From Eq. (i), we get $$ \begin{aligned} I_0+ & I_1+I_2+I_3+\ldots+I_7 \\ & =\tan x+\frac{\tan ^2 x}{2}+\frac{\tan ^5 x}{5}+\ldots . .+\frac{\tan ^6 x}{6} \end{aligned} $$ On comparing, we get $$ \begin{aligned} A_1 & =1, A_2=1 / 2, A_3=A_4=A_7=0, A_5=1 / 5 \\ A_6 & =1 / 6 \\ \sum_{k=1}^7 A_k & =A_1+A_2+A_3+\ldots+A_7 \\ & =1+\frac{1}{2}+\frac{1}{5}+\frac{1}{6}=\frac{56}{30}=\frac{28}{15} \end{aligned} $$
If $I_n=\int \tan ^n x d x \Rightarrow I_n+I_{n-2}=\frac{(\tan x)^{n-1}}{n-1}$ Put $n=2, I_0+I_2=\tan x \Rightarrow n=3, I_1+I_3=\frac{\tan ^2 x}{2}$ $$ n=6, I_4+I_6=\frac{\tan ^5 x}{5} \Rightarrow n=7, I_5+I_7=\frac{\tan ^6 x}{6} $$ From Eq. (i), we get $$ \begin{aligned} I_0+ & I_1+I_2+I_3+\ldots+I_7 \\ & =\tan x+\frac{\tan ^2 x}{2}+\frac{\tan ^5 x}{5}+\ldots . .+\frac{\tan ^6 x}{6} \end{aligned} $$ On comparing, we get $$ \begin{aligned} A_1 & =1, A_2=1 / 2, A_3=A_4=A_7=0, A_5=1 / 5 \\ A_6 & =1 / 6 \\ \sum_{k=1}^7 A_k & =A_1+A_2+A_3+\ldots+A_7 \\ & =1+\frac{1}{2}+\frac{1}{5}+\frac{1}{6}=\frac{56}{30}=\frac{28}{15} \end{aligned} $$
Hence, the correct answer is option: 'B'.