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Which is the best substitution to make to evaluate \int \frac{\cos(2x)}{\sqrt{5 - 2 \sin(2x)} dx?...

Question:

Which is the best substitution to make to evaluate {eq}\displaystyle \int \frac{\cos(2x)}{\sqrt{5 - 2 \sin(2x)}}\, dx {/eq}?

a. {eq}\; u = \sin(2x) {/eq}

b. {eq}\; u = \cos(2x) {/eq}

c. {eq}\; u = 2x {/eq}

d. {eq}\; u = 5 - 2 \sin(2x) {/eq}

Using Substitution to Solve Integral:


If we have a function of the form, {eq}f(g(x)) {/eq} then it's integral given by, {eq}\displaystyle \int f(g(x))dx {/eq} can be simplified using the substitution, {eq}t=g(x)\Rightarrow dt=g'(x)dx {/eq} and the integral simplifies to, {eq}\displaystyle \int f(t)\frac{dt}{h(t)} {/eq}

where {eq}h(t)=g'(x) {/eq} by using the substitution to convert {eq}g'(x) {/eq} into a function {eq}h(t) {/eq} of variable {eq}t {/eq}


Answer and Explanation:


The best strategy to identify the correct substitution includes,

1. Considering the available function which can be combined with the available {eq}\displaystyle dx {/eq} term.

2. The integral should be simplified by substituting for as large or complicated a function as possible.


In the given function to be integrated we have {eq}\displaystyle \sqrt{5 - 2 \sin(2x)} {/eq} as the denominator and {eq}\displaystyle \cos(2x) {/eq} as numerator , so we have {eq}\displaystyle \cos(2x) {/eq} available to be combined with {eq}\displaystyle dx {/eq} . Hence the first guess would be to substitute

$$\displaystyle t=\sin(2x) \Rightarrow dt=\cos(2x)(2)=2\cos(2x) $$.


But this will leave out the 2 multiplied and also the 5 from which it is subtracted after multiplication. So now we need to consider if we can substitute for a more complicated function getting almost the same derivative .i.e.

$$\displaystyle t=5-2\sin(2x)\Rightarrow dt=(0-2\cos(2x)(2)) \ dx=-4\cos(2x) \ dx $$


The only issue is about the 4 that is multiplied. We can take care of that by simply multiplying and dividing by 4 in the given function. So finally we can solve the integral using this substitution.


Tuus option d. is the best option. Also the integral will be,

{eq}\displaystyle \begin{align} &\int_{}^{} \frac{\cos(2x)}{\sqrt{5 - 2 \sin(2x)}} \ dx\\ &=\frac{-1}{4}\int_{}^{} \frac{-4\cos(2x)}{\sqrt{5 - 2 \sin(2x)}} \ dx\\ &=\frac{-1}{4}\int_{}^{}\frac{dt}{\sqrt{t}} &[\text{ Putting }t=5-2\sin(2x)\Rightarrow dt=(0-2\cos(2x)(2)) \ dx=-4\cos(2x) \ dx]\\ &=\frac{-1}{4}\left[ 2\sqrt{t}\right]\\ &=\frac{-\sqrt{t}}{2}\\ &=\frac{-\sqrt{5-2\sin(2x)}}{2}&[\text{ Re-substituting } t=5-2\sin(2x) ] \end{align} {/eq}


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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