Evaluate the integral. int(6sec^(2)(3t))/(7+2tan(3t))dt

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Evaluate the integral.   int(6sec^(2)(3t))/(7+2tan(3t))dt

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Identify Substitution: Let's start by identifying a substitution that can simplify the integral. We notice that the derivative of tan(3t) is sec^2(3t) multiplied by the constant 3. This suggests that a substitution involving tan(3t) might be useful. Let's set u = tan(3t), which means du/dt = 3sec^2(3t).Express in terms of u: Now we need to express the integral in terms of u. We have du = 3sec^2(3t)dt, so dt = du/(3sec^2(3t)). We also need to express the sec^2(3t) term in the numerator in terms of u. Since sec^2(3t) = 1 + tan^2(3t), we can write sec^2(3t) as 1 + u^2.Substitute u and dt: Substituting u and dt into the integral, we get: int(6sec^2(3t))/(7+2tan(3t))dt = int(6(1+u^2))/(7+2u) * (du/3(1+u^2)) Simplifying this, we get: int(6)/(7+2u) * (du/3) = int(2)/(7+2u) duIntegrate rational function: The integral int(2)/(7+2u) du is a simple rational function, and we can integrate it directly. The antiderivative of 1/(a+bu) with respect to u is (1/b)ln|a+bu|, so the antiderivative of 2/(7+2u) with respect to u is (1/2)ln|7+2u|.Substitute back to t: Now we substitute back the original variable t into our antiderivative. Since u = tan(3t), we have: (1/2)ln|7+2u| = (1/2)ln|7+2tan(3t)| + C, where C is the constant of integration.Final Antiderivative: We have found the antiderivative, so the integral of (6sec^2(3t))/(7+2tan(3t)) with respect to t is: (1/2)ln|7+2tan(3t)| + C

Evaluate the integral. $\newline$ $\int\frac{6\sec^{2}(3t)}{7+2\tan(3t)}\,dt$

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Evaluate the integral.\newline∫6sec⁡2(3t)7+2tan⁡(3t) dt\int\frac{6\sec^{2}(3t)}{7+2\tan(3t)}\,dt∫7+2tan(3t)6sec2(3t)​dt

Evaluate the integral. $\newline$ $\int\frac{6\sec^{2}(3t)}{7+2\tan(3t)}\,dt$