MCQ
$\int \sec ^2(7-4 x) d x=?$
  • A
    $-4 \tan (7-4 x)+C$
  • $\frac{-1}{4} \tan (7-4 x)+C$
  • C
    $4 \tan (7-4 x)+C$
  • D
    $\frac{1}{4} \tan (7-4 x)+C$

Answer

Correct option: B.
$\frac{-1}{4} \tan (7-4 x)+C$
Given integral is $\int \sec ^2(7-4 x) d x=$ ?
Let, $7-4 x = z$
$\Rightarrow-4 dx=dz$
So, $\int \sec ^2(7-4 x) dx=?$
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$\int \sec ^2(7-4 x) d x$ where $c$ is the integrating constant. 
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$=-\frac{1}{4} \tan z+c$
$=-\frac{1}{4} \tan (7-4 x)+c$

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