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Exercise 7.2 · Q22

Q.Integrate the following function: sec⁡2(7−4x)\sec^2 (7 - 4x)

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The key idea is to use the substitution u=7−4xu = 7 - 4x, which turns the integral into a standard form. The final answer is −14tan⁡(7−4x)+C-\frac{1}{4} \tan(7 - 4x) + C.

Why U-Substitution Works Here

When you see a function like sec⁡2\sec^2 wrapped around a linear expression like 7−4x7 - 4x, your first instinct should be: the derivative of the inside will simplify things. The derivative of 7−4x7 - 4x is −4-4, a constant. That constant can be pulled out of the integral, leaving you with a pure sec⁡2u\sec^2 u — which integrates directly to tan⁡u\tan u.

Without substitution, you'd be stuck trying to guess a composite antiderivative. With it, the problem becomes mechanical.

Step-by-Step Solution

  1. Set up the substitution. Let u=7−4xu = 7 - 4x. Then differentiate:

dudx=−4⇒du=−4 dx⇒dx=−14 du.\frac{du}{dx} = -4 \quad \Rightarrow \quad du = -4 \, dx \quad \Rightarrow \quad dx = -\frac{1}{4} \, du.

  1. Rewrite the integral in terms of uu. Replace 7−4x7 - 4x with uu, and dxdx with −14du-\frac{1}{4} du:

∫sec⁡2(7−4x) dx=∫sec⁡2(u)⋅(−14)du=−14∫sec⁡2u du.\int \sec^2(7 - 4x) \, dx = \int \sec^2(u) \cdot \left(-\frac{1}{4}\right) du = -\frac{1}{4} \int \sec^2 u \, du.

  1. Integrate the standard form. The antiderivative of sec⁡2u\sec^2 u is tan⁡u\tan u (this is a fundamental derivative fact: ddutan⁡u=sec⁡2u\frac{d}{du} \tan u = \sec^2 u). So:

−14∫sec⁡2u du=−14tan⁡u+C.-\frac{1}{4} \int \sec^2 u \, du = -\frac{1}{4} \tan u + C.

  1. Substitute back. Replace uu with 7−4x7 - 4x: …

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