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Chapter 4: Problem 37

Determine the following indefinite integrals. Check your work by differentiation. $$\int(\sin 2 y+\cos 3 y) d y$$

Short Answer

Expert verified
Question: Find the indefinite integral of the given function: $$\int(\sin 2 y+\cos 3 y) d y$$ Answer: $$\int(\sin 2 y+\cos 3 y) d y = -\frac{1}{2} \cos(2y) + \frac{1}{3} \sin(3y) + C$$

Step by step solution

01

Apply the integral of the sum rule

To find the indefinite integral of the given function, we will apply the integral of the sum rule which states that $$\int[f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx$$ In our case, we have $$\int(\sin 2 y+\cos 3 y) d y = \int \sin 2 y d y + \int \cos 3 y d y$$
02

Apply the trigonometric functions integral formulas

Now we need to find the indefinite integrals of both trigonometric functions. For the first integral, we have: $$\int \sin 2 y d y$$ Using the formula $$\int \sin(ax) dx = \frac{-1}{a} \cos(ax) + C$$, we get: $$-\frac{1}{2} \cos(2y) + C_1$$ For the second integral, we have: $$\int \cos 3 y d y$$ Using the formula $$\int \cos(ax) dx = \frac{1}{a} \sin(ax) + C$$, we get: $$\frac{1}{3} \sin(3y) + C_2$$ Now, we substitute both of these integrals back into our sum, we get: $$\int(\sin 2 y+\cos 3 y) d y = -\frac{1}{2} \cos(2y) + C_1 + \frac{1}{3} \sin(3y) + C_2$$ Combine the constants to get: $$\int(\sin 2 y+\cos 3 y) d y = -\frac{1}{2} \cos(2y) + \frac{1}{3} \sin(3y) + C$$
03

Check the result by differentiation

To verify our result, we must differentiate our result and compare it to the original function. Differentiating, we get: $$\frac{d}{dy} \left( -\frac{1}{2} \cos(2y) + \frac{1}{3} \sin(3y) + C \right) = (\sin 2 y+\cos 3 y)$$ We have shown that the indefinite integral is correct based on a comparison between the derived function and the original function. Therefore, our final answer is: $$\int(\sin 2 y+\cos 3 y) d y = -\frac{1}{2} \cos(2y) + \frac{1}{3} \sin(3y) + C$$

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