MCQ
$\int_{}^{} {{{\tan }^2}x\;dx} $ is equal to
- A$\tan x + x + c$
- ✓$\tan x - x + c$
- C$\sec x + x + c$
- D$\sec x - x + c$
✓
Answer
Correct option: B.
$\tan x - x + c$
b
(b)$\int_{}^{} {{{\tan }^2}x\,dx} = \int_{}^{} {({{\sec }^2}x - 1)\,dx} = \tan x - x + c$.
(b)$\int_{}^{} {{{\tan }^2}x\,dx} = \int_{}^{} {({{\sec }^2}x - 1)\,dx} = \tan x - x + c$.
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