Question 11 of 33intermediate🔧 ApplyShort Answer2 marks

Integrate the functions in Exercises 1 to 37: sec⁡2\sec^2(7 - 4x)

Correct Answer

−14tan⁡(7−4x)+C-\frac{1}{4} \tan(7 - 4x) + C

Exercise: EXERCISE 7.2 | Q: 22 | (Page 17)
For More Understanding

Explanation

This is a direct integration problem using substitution method. The context shows in point (6) that when x=atan⁡θx = a \tan \theta, we have dx=asec⁡2θdθdx = a \sec^2 \theta d\theta, which implies that the integral of sec⁡2θ\sec^2 \theta is tan⁡θ\tan \theta. Using this relationship with appropriate substitution, we can solve the given integral.

Solution Steps

  1. Step 1: Let t=7−4xt = 7 - 4x. Then dt=−4 dxdt = -4 \, dx which gives dx=−dt4dx = -\frac{dt}{4}

  2. Step 2: Substitute in the integral: ∫sec⁡2(7−4x) dx=∫sec⁡2(t)⋅(−dt4)=−14∫sec⁡2(t) dt\int \sec^2(7 - 4x) \, dx = \int \sec^2(t) \cdot \left(-\frac{dt}{4}\right) = -\frac{1}{4} \int \sec^2(t) \, dt

  3. Step 3: From the context, since dx=asec⁡2θdθdx = a \sec^2 \theta d\theta when x=atan⁡θx = a \tan \theta, we know ∫sec⁡2t dt=tan⁡t+C\int \sec^2 t \, dt = \tan t + C

  4. Step 4: Therefore, ∫sec⁡2(7−4x) dx=−14tan⁡(t)+C=−14tan⁡(7−4x)+C\int \sec^2(7 - 4x) \, dx = -\frac{1}{4} \tan(t) + C = -\frac{1}{4} \tan(7 - 4x) + C

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