Question

The indefinite integral $I=\int \frac{{\left(sec\right)}^{2}xtanx\left(secx+tanx\right)dx}{\left({\left(sec\right)}^{5}x+{\left(sec\right)}^{2}x{\left(tan\right)}^{3}x-{\left(sec\right)}^{3}x{\left(tan\right)}^{2}x-{\left(tan\right)}^{5}x\right)}$ simplifies to $\frac{1}{3}ln\left|f\left(x\right)\right|+c,$ where $f\left(\frac{\pi }{4}\right)=2\sqrt{2}+1$ and $c$ is the constant of integration. If the value of $f\left(\frac{\pi }{3}\right)$ is $a+\sqrt{b},$ then the value of $b-3a$ is equal to

Answer 1

Answer: 3

$I=\frac{1}{3}\int \frac{\left(3{\left(sec\right)}^{3}xtanx+3{\left(sec\right)}^{2}x{\left(tan\right)}^{2}x\right)dx}{\left({\left(sec\right)}^{2}x-{\left(tan\right)}^{2}x\right)\left({\left(sec\right)}^{3}x+{\left(tan\right)}^{3}x\right)}$
Let $se{c}^{3}x+ta{n}^{3}x=t$
Then, $I=\frac{1}{3}\int \frac{dt}{t}$
$=\frac{1}{3}ln\left|t\right|+c$
$=\frac{1}{3}ln\left|se{c}^{3}x+ta{n}^{3}x\right|+c$
$\therefore f\left(x\right)={\left(sec\right)}^{3}x+{\left(tan\right)}^{3}x$
$\Rightarrow f\left(\frac{\pi }{3}\right)=8+\sqrt{27}$
$\Rightarrow a=8,b=27$
$\Rightarrow b-3a=3$