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The integral between zero and π over two of cos π divided by the square root of sin π with respect to π is convergent.
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What does it converge to?
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Straight away, we can notice that the question has told us our integral is convergent.
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This gives us a hint that weβre looking at an improper integral.
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Now, since we donβt see any infinities as our limits of integration, itβs also implied that our integrand has a discontinuity on the closed interval between zero and π over two, which are the limits of the integral.
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Letβs look for this discontinuity.
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Since the integrand that weβve been given is a quotient, we should try and find where its denominator is equal to zero.
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The square root of sin π is zero, where sin π itself is also equal to zero.
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This occurs when π is equal to zero plus ππ, which is of course just ππ.
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And π here is an integer.
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So here we see that our denominator is equal to zero when π is equal to any of the following.
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The only one of these values which occurs over the interval of our integration is zero itself.
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Okay, for due diligence, letβs now check what happens to our numerator when π is equal to zero.
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Our numerator is cos π.
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So this becomes cos of zero, which is equal to one.
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We do this to check that when π equals zero, our function is not equal to the indeterminate form of zero over zero.
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Since we would then be looking at a removable discontinuity.
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In actual fact, when π is equal to zero, our function is equal to one over zero, which is the case of an infinite discontinuity.
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Okay.
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Weβve now confirmed that we have an improper integral.
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And we have an infinite discontinuity at the lower limit of integration.
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The definition of an improper integral specifically for this case tells us that if π is a continuous function on the interval which is open at π and closed at π and discontinuous at π.
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Then the integral between π and π of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches π from the right of the integral between π‘ and π of π of π₯ with respect to π₯.
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If this limit exists and is finite.
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Letβs now apply this to our question.
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The integral that weβve been given is equal to the limit as π‘ approaches zero from the right of the same integral, but now between the limits of π‘ and π over two.
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Now, at this point, we might notice that our integral is not entirely trivial.
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And weβll need to put in some work to evaluate it.
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To move forward, weβre gonna be using a π’-substitution, specifically π’ is equal to sin π.
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From this, we differentiate to find that dπ’ by dπ is equal to cos π.
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An equivalent statement to this is that dπ’ is equal to cos π dπ.
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We now do the following with our substitution.
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We replace sin π with π’.
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And we replace cos π dπ with dπ’.
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Weβre then left with the integral of one over the square root of π’ with respect to π’.
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Another way of expressing this integrand is π’ to the power of negative a half.
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Evaluating this, we get two π’ to the power of a half, which, is of course, two times the square root of π’.
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We can now go back to our original substitution and replace our π’ with sin π.
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We therefore have that our integral is equal to two times the square root of sin π, of course, plus the constant of integration.
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Now that weβve done the legwork on our integral, letβs use this result to move forward with our question.
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We use the antiderivative that weβve just found for our integral.
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And we get the following.
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We now input our limits of integration π‘ and π over two.
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And weβre left with the following limit.
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We can now take a direct substitution approach to our limit.
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And we can evaluate since sin of π over two is equal to one.
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And sin of zero is equal to zero.
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Weβre then left with two times the square of one minus two times the square root of zero.
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Of course, the square root of one is just one.
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And this entire term is just zero.
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The answer that weβre left with is therefore two.
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Upon reaching this result, we have answered our question.
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We used our definition to find that the integral given in the question converges to a value of two.