MCQ
$\int_0^{\pi /4} {{{\tan }^6}x \, {{\sec }^2}x\,dx = } $
- ✓$\frac{1}{7}$
- B$\frac{2}{7}$
- C$1$
- DNone of these
✓
Answer
Correct option: A.
$\frac{1}{7}$
a
(a) Put $t = \tan x \Rightarrow dt = {\sec ^2}x\,dx$
(a) Put $t = \tan x \Rightarrow dt = {\sec ^2}x\,dx$
Now $\int_0^{\pi /4} {{{\tan }^6}x{{\sec }^2}xdx = \int_0^1 {{t^6}dt = \frac{1}{7}[{t^7}]_0^1 = \frac{1}{7}} } $.
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