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Determine whether the integral from zero to β of the sin squared of πΌ with respect to πΌ is convergent or divergent.
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In this question, weβre given a definite integral, and weβre told to determine whether this definite integral is convergent or divergent.
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And the first thing we need to notice to answer this question is, this is an improper integral.
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This is because one of our limits of integration is positive or negative β.
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In this case, the upper limit of integration is β.
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So to answer this question, weβre first going to need to recall how do we evaluate an improper integral where the upper limit of integration is β.
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We recall the integral from π to β of π of π₯ with respect to π₯ is defined to be equal to the limit as π‘ approaches β of the integral from π to π‘ of π of π₯ with respect to π₯, where our values of π‘ must be bigger than π.
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When this limit converges, we say that our improper integral converges and is equal to the value of this limit.
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And when this limit does not converge, we say that our integral is divergent.
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So to determine the convergence or divergence of the integral given to us in the question, we need to construct this limit and determine the convergence or divergence of this limit.
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In our case, our integral is with respect to πΌ.
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So instead of π₯, weβre going to use the variable πΌ.
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Our lower limit of integration π is equal to zero, and our integrand is the sin squared of πΌ.
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So we set our function to be the sin squared.
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This gives us the integral from zero to β of the sin squared of πΌ with respect to πΌ is equal to the limit as π‘ approaches β of the integral from zero to π‘ of the sin squared of πΌ with respect to πΌ.
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And we need to determine the convergence or divergence of this limit.
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And to do this, weβre going to need to evaluate our definite integral.
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So weβre going to need to integrate the sin squared of πΌ.
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And the easiest way to do this is to use the double angle formula for cosine.
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Recall, the double angle formula for cosine tells us the cos of two πΌ will be equivalent to one minus two times the sin squared of πΌ.
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We want to rearrange this equivalence to make the sin squared of πΌ the subject.
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Adding two sin squared πΌ to both sides of our equivalence and subtracting cos of two πΌ from both sides, we get two sin squared of πΌ is equivalent to one minus the cos of two πΌ.
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Now, all we need to do is divide through by two.
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This gives us the sin squared of πΌ is equivalent to one minus the cos of two πΌ all divided by two.
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And we can see this is a much easier expression to integrate.
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So by using the double angle formula for cosine, we were able to rewrite this as the limit as π‘ approaches β of the integral from zero to π‘ of one minus the cos of two πΌ all divided by two with respect to πΌ.
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And weβre almost ready to evaluate this integral.
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However, instead of dividing this entire expression through by two, weβll divide each term individually by two.
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So we can rewrite our integrand as one-half minus the cos of two πΌ over two.
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Now, we can evaluate this integral term by term.
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First, we need to integrate the constant one-half with respect to πΌ.
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This is of course equal to πΌ over two.
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Next, we want to integrate negative the cos of two πΌ divided by two.
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And if we integrate negative the cosine function, we get negative the sine function.
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And remember, we need to divide by the coefficient of πΌ, which is two, giving us negative sin of two πΌ divided by four.
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And of course we need to evaluate this at the limits of integration πΌ is equal to zero and πΌ is equal to π‘.
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This gives us the following expression.
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The next thing we need to do is evaluate our antiderivative at the limits of integration.
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Doing this, we get the limit as π‘ approaches β of π‘ over two minus the sin of two π‘ over four minus zero over two minus the sin of two times zero over four.
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And we can simplify this.
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We see that zero over two and the sin of two times zero over four are both equal to zero.
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So this simplifies to just give us the limit as π‘ approaches β of π‘ over two minus the sin of two π‘ over four.
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And now weβre ready to evaluate this limit.
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As π‘ is approaching β, our first term π‘ over two is growing without bound.
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Itβs approaching β.
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However, as π‘ approaches β, the second term inside of our limit, the sin of two π‘ over four, is oscillating between negative one-quarter and one-quarter.
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It doesnβt converge to any value.
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However, we can conclude something about this limit.
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The first term is growing without bound, and the second term is bounded.
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This means our entire limit is growing without bound.
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Therefore, this limit is divergent.
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And remember, this limit being divergent means that our improper integral is also divergent.
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Therefore, we were able to show the integral from zero to β of the sin squared of πΌ with respect to πΌ is divergent.