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Question

Question: $\int \sec^2(2-3x)dx$...

∫sec⁡2(2−3x)dx\int \sec^2(2-3x)dx

Answer

−13tan⁡(2−3x)+C-\frac{1}{3} \tan(2-3x) + C

Explanation

Solution

Let u=2−3xu = 2-3x, so du=−3dxdu = -3dx, which means dx=−13dudx = -\frac{1}{3}du. The integral becomes −13∫sec⁡2(u)du=−13tan⁡(u)+C=−13tan⁡(2−3x)+C-\frac{1}{3} \int \sec^2(u) du = -\frac{1}{3} \tan(u) + C = -\frac{1}{3} \tan(2-3x) + C