Question

If $\int \left((\sqrt{\tan x}{)}^{17}-\sqrt{\tan x}\right)dx=2(\tan x{)}^{\frac{3}{2}}\left(\frac{1}{a}(\tan x{)}^6+\frac{1}{b}(\tan x{)}^4+\frac{1}{c}(\tan x{)}^2+\frac{1}{d}\right)+M$ 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 $\tan x=t^2\Rightarrow {\sec }^{2}xdx=2tdt\Rightarrow dx=\left(\frac{2t}{1+t^4}\right)dt$
$\int t\left(t^{16}-1\right)\left(\frac{2t}{1+t^4}\right)dt=2\int t^2\left(t^8+1\right)\left(t^4-1\right)dt=2\int \left(t^{14}-t^{10}+t^6-t^2\right)dt$
$=2\left(\frac{t^{15}}{15}-\frac{t^{11}}{11}+\frac{t^7}{7}-\frac{t^3}{3}\right)+M=2\cdot t^3\left(\frac{t^{12}}{15}-\frac{t^8}{11}+\frac{t^4}{7}-\frac{1}{3}\right)+M$
$=2(\tan x{)}^{\frac{3}{2}}\left(\frac{1}{15}(\tan x{)}^6+\frac{1}{(-11)}(\tan x{)}^4+\frac{1}{7}(\tan x{)}^2+\frac{1}{(-3)}\right)+M$
$\therefore \text{ Hence, }a+b+c+d=15-11+7-3=8$