MCQ
Evaluate: $\int \sec ^2(7-4 x) d x$
- A$\frac{1}{4} \tan (7-4 x)+C$
- B$\frac{1}{4} \tan (7+4 x)+C$
- C$\frac{-1}{4} \tan (7+4 x)+C$
- ✓$\frac{-1}{4} \tan (7-4 x)+C$
✓
Answer
Correct option: D.
$\frac{-1}{4} \tan (7-4 x)+C$
(d) : Let $I=\int \sec ^2(7-4 x) d x$
Put $7-4 x=t \Rightarrow d x=\frac{-1}{4} d t$
$
\therefore \quad I=\int \frac{\sec ^2 t}{-4} d t \Rightarrow I=\frac{\tan t}{-4}+C=\frac{\tan (7-4 x)}{-4}+C
$
Put $7-4 x=t \Rightarrow d x=\frac{-1}{4} d t$
$
\therefore \quad I=\int \frac{\sec ^2 t}{-4} d t \Rightarrow I=\frac{\tan t}{-4}+C=\frac{\tan (7-4 x)}{-4}+C
$
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