Question

${\displaystyle \int \sqrt{\tan x} {\sec }^4 x dx}$

Answer 1

${\displaystyle \int \sqrt{\tan x} {\sec }^4 x dx}$
$$=\int \sqrt{\tan x}\cdot {\sec }^{2}x\cdot {\sec }^{2}x\text{ dx }$$
$$=\int \sqrt{\tan x}\cdot (1+{\tan }^{2}x){\sec }^{2}x\text{ dx }$$
$$\text{Let }\tan x=t$$
$$\Rightarrow {\sec }^{2}x\text{ dx }=dt$$
$$Now,\int \sqrt{\tan x}\cdot (1+{\tan }^{2}x){\sec }^{2}x\text{ dx }$$
$$=\int \sqrt{t}(1+t^2)dt$$
$$=\int (\sqrt{t}+t^{\frac{5}{2}})dt$$
$$=\int (t^{\frac{1}{2}}+t^{\frac{5}{2}})dt$$
$$=\frac{2}{3}{t}^{\frac{3}{2}}+\frac{2}{7}{t}^{\frac{7}{2}}+C$$
$$=\frac{2}{3}{\tan }^{\frac{3}{2}}x+\frac{2}{7}{\tan }^{\frac{7}{2}}x+C$$