Question
Evaluate the following integrals:
$\int\sqrt{\tan\text{x}}\sec^4\text{x}\text{ dx}$

Answer

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

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