Question
Evaluate the following integrals:
$\int\tan\text{x }\sec^4\text{x}\text{dx}$
$\int\tan\text{x }\sec^4\text{x}\text{dx}$
✓
Answer
Let $\text{I}=\int\tan\text{x }\sec^4\text{x}\text{dx}$ Then
$\text{I}=\int\tan\text{x }\sec^2\text{x}\sec^2\text{x}\text{dx}$
$=\int\tan\text{x}(1+\tan^2\text{x})\sec^2\text{x}\text{dx}$
$\text{I}=\int\big(\tan\text{x}+\tan^3\text{x}\big)\sec^2\text{x}\text{dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{xdx}=\text{dt},$ we get
$\text{I}=\int(\text{t}+\text{t}^3)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4}{4}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\times\tan^2\text{x}+\frac{1}{4}\times\tan^4\text{x}+\text{C}$
$\text{I}=\int\tan\text{x }\sec^2\text{x}\sec^2\text{x}\text{dx}$
$=\int\tan\text{x}(1+\tan^2\text{x})\sec^2\text{x}\text{dx}$
$\text{I}=\int\big(\tan\text{x}+\tan^3\text{x}\big)\sec^2\text{x}\text{dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{xdx}=\text{dt},$ we get
$\text{I}=\int(\text{t}+\text{t}^3)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4}{4}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\times\tan^2\text{x}+\frac{1}{4}\times\tan^4\text{x}+\text{C}$
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