Understanding Integrals: Techniques and Solutions in Trig Identities

[shelovesmath]

I'm doing a chapter on general techniques of integration, and I have a complete solutions manual (this is number 37 in stewards early transcendentals 6e chapter 7 review) but I'm not following the method from one step to the next.

Integral: (1+ sin2x) cos2x dx becomes integral:cos2x + 1/2 integral: sin4x

I can see they obviously distributed and then split one interval into two.

I know it's trig identity stuff, but I'm totally lost on this one. Any help is much appreciated!

 

[cepheid]

You might recall a set identities called the sum and difference formulae that tell you the sine or cosine of the sum or difference of two angles. As it happens, for sine:

$$\sin(\theta \pm \phi) = \sin\theta\cos\phi \pm \cos\theta\sin\phi$$​

Now, what happens if θ = φ, and if we're adding the two angles together? Then:

$$\sin(\theta + \theta) = \sin(2\theta) = \sin\theta\cos\theta + \cos\theta\sin\theta$$

$$= 2\sin\theta\cos\theta$$
​

This leads to the result that

$$\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)$$​

Now, we can apply this result to your integral, with θ = 2x:

$$\int [1+ \sin(2x)]\cos(2x)\,dx = \int [\cos(2x) + \sin(2x)\cos(2x)]\,dx$$​

Using the identity we derived above on the second term in the integrand, we obtain:

$$= \int \cos(2x)\,dx + \frac{1}{2}\int \sin(4x)\,dx$$​

From here, I think you can solve it to get the solution in the manual.

 

[shelovesmath]

Thanks!