Question
Evaluate the following integrals:
$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
✓
Answer
$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$\text{Let, }\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$=\int\text{t}^\frac{3}{2}\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+\text{C}$
$=\frac{2}{5}\tan^\frac{5}{2}\text{x}+\text{C}$
$\text{Let, }\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$=\int\text{t}^\frac{3}{2}\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+\text{C}$
$=\frac{2}{5}\tan^\frac{5}{2}\text{x}+\text{C}$
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