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7. (a) Find \( \int(3\cos2x + 1) \, dx \) - Scottish Highers Maths - Question 7 - 2015
Question 7
7. (a) Find \( \int(3\cos2x + 1) \, dx \). (b) Show that \( 3\cos2x + 1 = 4\cos^2x - 2\sin^2x \). (c) Hence, or otherwise, find \( \int[\sin^2x - 2\cos^2 x] \, dx ... show full transcript
Worked Solution & Example Answer:7. (a) Find \( \int(3\cos2x + 1) \, dx \) - Scottish Highers Maths - Question 7 - 2015
Step 1
(a) Find \( \int(3\cos2x + 1) \, dx \)
Answer
To find ( \int(3\cos2x + 1) , dx ), we can separate the integral into two parts:
Starting with the first integral: [ \int 3\cos2x , dx = 3 \cdot \frac{1}{2} \sin2x + C = \frac{3}{2}\sin2x + C. ]
Now, the second integral: [ \int 1 , dx = x + C. ]
Combining both results, we have: [ \int(3\cos2x + 1) , dx = \frac{3}{2}\sin2x + x + C. ]
Step 2
(b) Show that \( 3\cos2x + 1 = 4\cos^2x - 2\sin^2x \)
Answer
To demonstrate the identity, we start by recalling the double angle formulas: [ \cos2x = 2\cos^2x - 1 \text{ and } \sin^2x = 1 - \cos^2x. ]
Substituting ( \cos2x ) into the expression: [ 3\cos2x + 1 = 3(2\cos^2x - 1) + 1 = 6\cos^2x - 3 + 1 = 6\cos^2x - 2. ]
Now, we recognize that: [ 2\sin^2x = 2(1 - \cos^2x) = 2 - 2\cos^2x. ]
Thus: [ 4\cos^2x - 2\sin^2x = 4\cos^2x - (2 - 2\cos^2x) = 6\cos^2x - 2. ]
This proves that: [ 3\cos2x + 1 = 4\cos^2x - 2\sin^2x. ]
Step 3
(c) Hence, or otherwise, find \( \int[\sin^2x - 2\cos^2 x] \, dx \)
Answer
To solve ( \int[\sin^2x - 2\cos^2 x] , dx ), we can rewrite the integrand using the identity ( \sin^2x = 1 - \cos^2x ):
[ \sin^2x - 2\cos^2x = (1 - \cos^2x) - 2\cos^2x = 1 - 3\cos^2x. ]
Now we can separate the integral: [ \int[1 - 3\cos^2x] , dx = \int 1 , dx - 3 \int \cos^2x , dx. ]
Using the identity ( \int \cos^2x , dx = \frac{x}{2} + \frac{1}{4} \sin2x + C ): [ \int 1 , dx = x, ] [ \int \cos^2x , dx = \frac{x}{2} + \frac{1}{4} \sin2x + C. ]
Thus: [ \int[\sin^2x - 2\cos^2x] , dx = x - 3\left( \frac{x}{2} + \frac{1}{4}\sin2x \right) + C = x - \frac{3x}{2} - \frac{3}{4}\sin2x + C = -\frac{x}{2} - \frac{3}{4}\sin2x + C. ]