MCQ
$\int_{}^{} {{{\sec }^p}x\tan x\;dx = } $
- A$\frac{{{{\sec }^{p + 1}}x}}{{p + 1}} + c$
- ✓$\frac{{{{\sec }^p}x}}{p} + c$
- C$\frac{{{{\tan }^{p + 1}}x}}{{p + 1}} + c$
- D$\frac{{{{\tan }^p}x}}{p} + c$
therefore $\int_{}^{} {{{\sec }^p}x\tan x\,dx} = \int_{}^{} {{t^{p - 1}}dt = \frac{{{t^p}}}{p} + c} = \frac{{{{\sec }^p}x}}{p} + c.$
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(where $p$ is a constant)
$f(x)= \begin{cases}n(1-2 n x) \text { if } 0 \leq x \leq \frac{1}{2 n}2 n(2 n x-1) \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n}4 n(1-n x) \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(n x-1) \text { if } \frac{1}{n} \leq x \leq 1\end{cases}$
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is $4$ , then the maximum value of the function $f$ is