Question

Integrate $\frac{{\tan }^4\sqrt{x}+{\sec }^2\sqrt{x}}{\sqrt{x}}$

The solution is $\frac{2{\tan }^3\left(\sqrt{x}\right)}{m}-2\tan \sqrt{x}+2\sqrt{x}+2\tan \sqrt{x}+C$.Find m

Answer 1

$let\sqrt{x}=tthen\left(\frac{1}{2\sqrt{x}}\right)dx=dt\therefore I=2\int (ta{n}^{4}t+se{c}^{2}t)dt=2\int \left[ta{n}^{2}t\right(se{c}^{2}t-1)+se{c}^{2}t]dt=2\int ta{n}^{2}tse{c}^{2}tdt-2\int ta{n}^{2}tdt+2\int se{c}^{2}tdt=2\left(\frac{ta{n}^{3}t}{3}\right)-2\int (se{c}^{2}t-1)dt+2tant+c=\left(\frac{2}{3}\right)ta{n}^{3}t-2tant+2t+2tant+c=\left(\frac{2}{3}\right)ta{n}^{3}t-4tant+2t+c=\left(\frac{2}{3}\right)ta{n}^{3}t\left(\sqrt{x}\right)-2tan\sqrt{x}+2tan\sqrt{x}+2\sqrt{x}+c\therefore m=3$