MCQ
$\int_0^{\pi / 4} \tan ^2 x d x=$
- ✓$1-\frac{\pi}{4}$
- B$1+\frac{\pi}{4}$
- C$\frac{\pi}{4}-1$
- D$\frac{\pi}{4}$
✓
Answer
Correct option: A.
$1-\frac{\pi}{4}$
(A)
$\int_0^{\pi / 4} \tan ^2 x d x=\int_0^{\pi / 4}\left(\sec ^2 x-1\right) d x$
$\begin{array}{l}=[\tan x]_0^{\frac{\pi}{4}}-[x]_0^{\frac{\pi}{4}} \\ =1-\frac{\pi}{4}\end{array}$
$\int_0^{\pi / 4} \tan ^2 x d x=\int_0^{\pi / 4}\left(\sec ^2 x-1\right) d x$
$\begin{array}{l}=[\tan x]_0^{\frac{\pi}{4}}-[x]_0^{\frac{\pi}{4}} \\ =1-\frac{\pi}{4}\end{array}$
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