Integrate the function sec ^{2}(7-4 x). | vedclass

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(A) Let $7-4x = t$. Then,differentiating both sides with respect to $x$,we get $-4 dx = dt$,which implies $dx = -\frac{1}{4} dt$. Substituting these i

Vedclass · Accepted Answer

$-\frac{1}{4} 	an(7-4x) + C$

Vedclass · Answer

(A) Let $7-4x = t$. Then,differentiating both sides with respect to $x$,we get $-4 dx = dt$,which implies $dx = -\frac{1}{4} dt$. Substituting these into the integral: $\int \sec^{2}(7-4x) dx = \int \sec^{2}(t) \left(-\frac{1}{4}ight) dt$ $= -\frac{1}{4} \int \sec^{2}(t) dt$ $= -\frac{1}{4} 	an(t) + C$ $= -\frac{1}{4} 	an(7-4x) + C$,where $C$ is an arbitrary constant.

Integrate the function $\sec ^{2}(7-4 x)$.

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