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In this video, weβll learn how we can use the antiderivative of inverse trigonometric functions to integrate more complicated functions where itβs not immediately obvious how the substitution rule and integration by parts can help.
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Itβs, therefore, important your confident in differentiating inverse trigonometric functions of the form inverse sine of π₯, inverse tan of π₯, and inverse secant π₯.
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This entire video is made possible thanks to the fundamental theorem of calculus.
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Remember, the first part of this theorem says that if a function π is continuous, then the derivative of the integral of a function π with respect to the variable π‘ over some interval π to π₯ is equal to the function π with respect to π₯.
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Essentially, it describes the derivative and integral as inverse processes.
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And luckily for us, this means that if we can recognize the derivative of a function as our integrand, with a little manipulation, we can quite easily integrate fairly nasty functions.
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So at this stage, weβre going to recall the general results for the derivatives of the key inverse trigonometric functions that weβll be referring to throughout this video.
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The first weβre interested in is the derivative of the function inverse sin of π₯ over π for real constants π.
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Itβs one over the square root of π squared minus π₯ squared.
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And this is only valid for values of inverse sin of π₯ over π greater than or equal to negative π by two and less than π by two.
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With similar constraints on the range of the function inverse tan of π₯ over π, we obtain its derivative to be π over π squared plus π₯ squared.
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And finally, we know that the derivative of the inverse sec of π₯ over π is π over π₯ times the square root of π₯ squared minus π squared.
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Now remember, when performing calculus with trigonometric functions, we always work with radians.
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Weβre now going to have look at a simple example of how these derivatives can help us to evaluate a function with a simple inverse trigonometric function as its result.
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Evaluate the definite integral between the limits of one and root three of negative one over one plus π₯ squared with respect to π₯.
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Here, we have a definite integral with limits of one and a square root of three.
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This means weβre going to need to use the second part of the fundamental theorem of calculus to evaluate it.
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This tells us that if π is a real value function on some closed interval π to π and capital πΉ is an antiderivative of π in that closed interval such that capital πΉ prime of π₯ is equal to π of π₯.
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Then if π is Riemann integrable on the closed interval, then we can say that the definite integral between π and π of π of π₯ is equal to capital πΉ of π minus capital πΉ of π.
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Essentially, we can evaluate the integral here by finding the antiderivative of this function negative one over one plus π₯ squared and evaluating it between root three and one.
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Now, the integral of negative one over one plus π₯ squared is not particularly nice.
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But actually we spot that we can take out any constant factors and focus on the integral itself.
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So here, we take our constant factor of negative one.
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And weβre now looking to evaluate the negative integral of one over one plus π₯ squared between one and root three.
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Then we spot that we have the standard result for the derivative of the inverse tan of π₯ over π.
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Itβs π over π squared plus π₯ squared.
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And this, of course, means that the antiderivative of π over π squared plus π₯ squared must be the inverse tan of π₯ over π.
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Now, if we compare the function π over π squared plus π₯ squared with our function one over one plus π₯ squared, we can see that π is equal to one.
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So if we say that π of π₯ is equal to one over one plus π₯ squared, then the antiderivative capital πΉ of π₯ must be the inverse tan of π₯ over one, which is simply the inverse tan of π₯.
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By the second part of our theorem then, we can say that the definite integral between one and root three of one over one plus π₯ squared with respect to π₯ is equal to the inverse tan of root three minus the inverse tan of one.
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And of course, we took out that constant of negative one at the start.
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We then know that the inverse tan of root three is π by three, and the inverse tan of one is π by four.
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Weβre going to find the difference between these two fractions by using a common denominator.
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We multiply the numerator and denominator of our first fraction by four and the numerator and denominator of our second fraction by three.
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And weβre looking to work out negative four π over 12 minus three π over 12.
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Four π by 12 minus three π by 12 is π by 12.
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So our answer here is negative π by 12.
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Now, we have just found the general result for the indefinite integral of π over π squared plus π₯ squared with respect π₯ for real constants π.
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Itβs the inverse tan of π₯ over π plus some constant of integration π.
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And rather than jumping straight into evaluating capital πΉ of π minus capital πΉ of π, we could, of course, have included this extra step using those square brackets.
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Find the most general antiderivative capital πΊ of π£ of the function π of π£ equals four cos π£ plus three over five root one minus π£ squared.
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Remember, the antiderivative is basically the opposite of the derivative.
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And another way of thinking about it is to find the antiderivative capital πΊ of π£.
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We can find the indefinite integral of this function.
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We, therefore, say that capital πΊ of π£ is equal to the indefinite integral of lower case π of π£.
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Letβs replace π of π£ with the function four cos π£ plus three over five times the square root of one minus π£ squared.
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We then recall a key property of integrals; that is, the integral of the sum of two or more functions is equal to the sum of the integral of each respective function.
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And we can, therefore, split our integral up.
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And we see that capital πΊ of π£ is equal to the integral of four cos π£ plus the integral of three over five times the square root of one minus π£ squared.
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Another key property we can apply is that the integral of some constant times a function is equal to that constant times the integral of the function.
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And so we can further rewrite this as four times the integral of cos of π£ plus three-fifths times the integral of one over the square of one minus π£ squared.
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Now, this is great because we can use in general results for derivatives.
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Firstly, we know that the derivative of sin π₯ is cos of π₯.
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So the antiderivative and, therefore, the integral of cos of π£ is sin of π£.
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And of course, when dealing with definite integrals, we add a constant of integration.
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Letβs call that π΄.
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So this first part becomes four times sin of π£ plus π΄.
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Next, we know that if the inverse sin of π₯ over π is greater than or equal to negative π by two and less than or equal to π by two, then its derivative is equal to one over the square root of π squared minus π₯ squared.
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Now, in our example, π squared is equal to one.
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So π must be equal to one also.
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So the antiderivative of one over the square root of one minus π£ squared and, therefore, the integral of this function is the inverse sin of π₯ over one.
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And we add another constant of integration π΅.
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Now, of course, the inverse sin of π₯ over one can be written as the inverse sin of π₯.
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Distributing our parentheses and combining the constants into one new constant capital πΆ, we find that the general antiderivative capital πΊ of π£ is equal to four sin of π£ plus three-fifths of the inverse sin of π£ plus πΆ.
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And in this example, weβve seen that the indefinite integral of one over the square root of π squared minus π₯ squared with respect to π₯ is the inverse sin of π₯ over π plus π.
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Weβre now going to look at an example that requires just a little more manipulation.
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Evaluate the indefinite integral of one over the square root of four π₯ squared minus 16 with respect to π₯.
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At first glance, it might appear that this has a simple result of an inverse trigonometric function.
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However, once we know that the derivative of the inverse sec of π₯ over π is π over π₯ times the square root of π₯ squared minus π squared, when the inverse secant of π₯ over π is greater than zero, less than π, but not equal to π by two.
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Our integrand isnβt quite of this form.
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We noticed particularly that we have four π₯ squared instead of just π₯ squared.
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Weβre, therefore, going to need to perform some manipulation.
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Weβre going to begin by multiplying both the numerator and denominator of our fraction by two.
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Now remember, this doesnβt actually change the integrand because itβs the equivalent of multiplying by one.
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So weβre looking to evaluate the indefinite integral of two over two π₯ times the square of four π₯ squared minus 16.
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Now, we notice that four π₯ squared and 16 are both square numbers.
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This means we can write four π₯ squared minus 16 as two π₯ all squared minus four squared.
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And now, we notice that we can make a substitution.
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If we let π’ be equal to two π₯ squared, then inside our square root sign, weβll have π’ squared minus four squared.
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Notice that thatβs a lot closer to the form weβre looking for.
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If π’ was equal to two π₯, then we know that dπ’ by dπ₯ must be equal to two.
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And we can say equivalently that dπ’ must be equal to two π₯.
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Well notice, we can now replace two dπ₯ with dπ’.
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We can replace two π₯ and two π₯ with π’, and our function is now in the form weβre looking for.
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We need to integrate one over π’ times the square root of π’ squared minus four squared with respect to π’.
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Looking back to our original derivative, we noticed that the inverse sec of π₯ over π is the antiderivative of π over π₯ times the square root of π₯ squared minus π squared.
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In our example, π must be equal to four.
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Now, the numerator, of course, of our fraction is one not four.
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So our integral will be a quarter of the inverse sec of π’ over four plus π.
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Our final step is a look back at our substitution, and we replace π’ with two π₯.
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And we see that our indefinite integral is a quarter of the inverse sec of π₯ over two plus the constant of integration π.
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In our final example, weβre going to look at an integral which involves manipulation by completing the square and a clever little substitution.
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Evaluate the indefinite integral of one over π₯ squared minus π₯ plus one with respect to π₯.
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Now, this isnβt a nice function to integrate at all.
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So weβre going to need to do something a little bit clever.
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Itβs certainly not the product of two functions.
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So weβre not going to use integration by parts.
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But if we do something special to the denominator, we can actually use integration by substitution.
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Weβre going to complete the square of the denominator of the expression π₯ squared minus π₯ plus one.
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Remember, we halve the coefficient of π₯.
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Here, thatβs negative one, so half of that is negative one-half.
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We, therefore, have π₯ minus a half all squared in the brackets.
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Negative a half squared is one-quarter, so we subtract that one quarter.
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And we see that our expression is equivalent to π₯ minus a half all squared plus three quarters.
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And now, this is the integral that weβre looking to evaluate.
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Next, we need to spot that we know that indefinite integral of π over π squared plus π₯ squared.
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Itβs the inverse tan of π₯ over π.
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So to ensure that our function looks a little like this, weβre going to perform a substitution.
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Weβre going to let π₯ minus a half be equal to π’.
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Then this part will be π’ squared.
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The derivative of π₯ minus one-half is one.
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So dπ’ by dπ₯ equals one, which means that dπ’ is equal to dπ₯.
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So we can replace dπ₯ with dπ’ and π₯ minus a half with π’.
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And we see that weβre actually looking to find the indefinite integral of one over π’ squared plus three-quarters.
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Now, this still doesnβt quite look like what weβre after.
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We need it to be π squared on the denominator.
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Well, three-quarters is the same as the square root of three quarters squared.
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So π here is equal to the square root of three-quarters.
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And of course, since the numerates of our fraction is one and not the square root of three-quarters, the integral is one divided by the square root of three-quarters times the inverse tan of π’ over the square root of three quarters plus π.
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One divided by the square root of three quarters is two root three over three.
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And then we go back to our substitution π’ equals π₯ minus a half.
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And we replace that in our result.
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And finally, we distribute our parentheses.
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The indefinite integral of one over π₯ squared minus π₯ plus one with respect to π₯ is two root three over three times the inverse tan of root three over three times two π₯ minus one plus the constant of integration π.
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In this video, weβve seen that we can use the concept of antiderivatives to integrate functions which have inverse trigonometric results.
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The integral of one over the square root of π squared minus π₯ squared is the inverse sin of π₯ over π.
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The integral of π over π squared plus π₯ squared is the inverse tan of π₯ over π.
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And the integral of π over π₯ times the square root of π₯ squared minus π squared is the inverse sec of π₯ over π.
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We also saw that sometimes we need to perform some manipulation and a clever substitution to achieve our results.