Question
Evaluate the following integrals:
$\int\tan\text{x}\sec^2\text{x}\sqrt{1-\tan^2\text{x}}\text{ dx}$
$\int\tan\text{x}\sec^2\text{x}\sqrt{1-\tan^2\text{x}}\text{ dx}$
Putting
$1-\tan^2\text{x}=\text{t}$ and $\tan\text{x}\sec^2\text{x}\text{ dx}=-\frac{\text{dt}}{2}$ in equation (1),We get
$\text{I}=\int\sqrt{\text{t}}\times\frac{-\text{dt}}{2}$
$=\frac{-1}{2}\int\text{t}^{\frac{1}{2}}\text{dt}$
$=-\frac{1}{2}\times\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=-\frac{1}{3}\text{t}^\frac{3}{2}+\text{C}$
$=-\frac{1}{3}\big[1-\tan^2\text{x}\big]^{\frac{3}{2}}+\text{C}$
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is continuous at X = 0.
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