Question

If $\int\left((\sqrt{\tan x})^{17}-\sqrt{\tan x}\right) d x=2(\tan x)^{\frac{3}{2}}\left(\frac{1}{a}(\tan x)^6+\frac{1}{b}(\tan x)^4+\frac{1}{c}(\tan x)^2+\frac{1}{d}\right)+M$ where $a , b , c$ and $d$ are real constants and $M$ is the constant of integration then $( a + b + c + d )$ equals

• A. 46
• B. 36
• C. 8
• D. 6

Answer 1

Answer: (C)

$\int \sqrt{\tan x}\left(\tan ^8 x-1\right) dx$
Put $\tan x=t^2 \Rightarrow \sec ^2 x d x=2 t d t \Rightarrow d x=\left(\frac{2 t}{1+t^4}\right) d t$
$\int t \left( t ^{16}-1\right)\left(\frac{2 t }{1+ t ^4}\right) dt =2 \int t ^2\left( t ^8+1\right)\left( t ^4-1\right) dt =2 \int\left( t ^{14}- t ^{10}+ t ^6- t ^2\right) dt $
$=2\left(\frac{ t ^{15}}{15}-\frac{ t ^{11}}{11}+\frac{ t ^7}{7}-\frac{ t ^3}{3}\right)+ M =2 \cdot t ^3\left(\frac{ t ^{12}}{15}-\frac{ t ^8}{11}+\frac{ t ^4}{7}-\frac{1}{3}\right)+ M$
$=2(\tan x )^{\frac{3}{2}}\left(\frac{1}{15}(\tan x )^6+\frac{1}{(-11)}(\tan x )^4+\frac{1}{7}(\tan x )^2+\frac{1}{(-3)}\right)+ M $
$\therefore \text { Hence, } a + b + c + d =15-11+7-3=8$