int tan(3x - 5) sec(3x - 5) , dx = | vedclass

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Question

(B) Let $t = 3x - 5$. Then,$dt = 3 \, dx$,which implies $dx = \frac{1}{3} \, dt$. Substituting these into the integral: $\int 	an(3x - 5) \sec(3x - 5

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$\frac{1}{3} \sec(3x - 5) + c$

Vedclass · Answer

(B) Let $t = 3x - 5$. Then,$dt = 3 \, dx$,which implies $dx = \frac{1}{3} \, dt$. Substituting these into the integral: $\int 	an(3x - 5) \sec(3x - 5) \, dx = \int 	an(t) \sec(t) \cdot \frac{1}{3} \, dt$ $= \frac{1}{3} \int \sec(t) 	an(t) \, dt$ Since the integral of $\sec(t) 	an(t)$ is $\sec(t) + c$,we get: $= \frac{1}{3} \sec(t) + c$ Substituting back $t = 3x - 5$: $= \frac{1}{3} \sec(3x - 5) + c$.

$\int \tan(3x - 5) \sec(3x - 5) \, dx = $

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