Question

The indefinite integral $I=\displaystyle \int \frac{\left(sec\right)^{2} x tan ⁡ x \left(sec ⁡ x + tan ⁡ x\right) d x}{\left(\left(sec\right)^{5} ⁡ x + \left(sec\right)^{2} ⁡ x \left(tan\right)^{3} ⁡ x - \left(sec\right)^{3} ⁡ x \left(tan\right)^{2} ⁡ x - \left(tan\right)^{5} ⁡ x\right)}$ 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

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