Question

$\int \tan x\sqrt{1-{\tan }^{2}x}{\sec }^{2}xdx=$

• A. $\frac{-1}{3}(1-{\tan }^{2}x)\sqrt{1-{\tan }^{2}x}+c$
• B. $(1-{\tan }^{2}x{)}^{3/2}+c$
• C. $\frac{1}{3}(1-{\tan }^{2}x{)}^{3/2}+c$
• D. $-(1-{\tan }^{2}x{)}^{3/2}+c$

Answer 1

Answer: (A)

$\frac{-1}{3}(1-{\tan }^{2}x)\sqrt{1-{\tan }^{2}x}+c$

$I=\int \tan x\sqrt{1-{\tan }^{2}x}{\sec }^{2}xdx$

Put $1-{\tan }^{2}x=t$

$-2\tan x{\sec }^{2}xdx=dt$

$I=-\frac{1}{2}\int \sqrt{t}dt$

$I=-\frac{1}{3}{t}^{3/2}+c$

$I=-\frac{1}{3}{(1-{\tan }^{2}x)}^{3/2}+c$