# Question Video: Finding the Point of Inflection of a Function Involving Using the Product Rule with Logarithmic Functions If Any Mathematics • Higher Education

Evaluate the integral β«_(0)^(π/4) (sec π‘ tan π‘ π’ + π‘ cos 2π‘ π£ + sinΒ² 2π‘ cos 2π‘ π€) dπ‘.

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### Video Transcript

Evaluate the definite integral between zero and π by four of sec of π‘ tan of π‘ π’ plus π‘ times cos of two π‘ π£ plus sin squared of two π‘ times cos of two π‘ π€ with respect to π‘.

Remember, to integrate a vector-valued function, we simply integrate each component in the usual way. Since this is a definite integral, we can use the fundamental theorem of calculus, which says that the definite integral between π and π of π« of π‘ with respect to π‘ is equal to the antiderivative capital π evaluated at π minus capital π evaluated at π. So essentially, all weβre going to do is integrate each of our components with respect to π‘ and evaluate them individually between the limits of zero and π by four.

Letβs complete this process for the component for π’. Itβs the definite integral between zero and π by four of sec π‘ tan π‘. We might recall the derivative of sec of π‘ is equal to sec π‘ tan π‘. So this means the antiderivative of sec π‘ tan π‘ must be sec π‘. So we use the fundamental theorem of calculus. And we see that this is equal to sec of π by four minus sec of zero. But of course, sec of π‘ is equal to one over cos of π‘. So we need to work out one over cos of π by four minus one over cos of zero. Thatβs root two minus one.

Weβve cleared some space so we can evaluate the definite integral between zero and π by four of π‘ times cos of two π‘. This time, we have the product of two functions. So weβre going to use integration by parts to evaluate. This says that the integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the integral of π£ times dπ’ by dπ₯ with respect to π₯. Weβll let π’ be equal to π‘. And thatβs because we know that dπ’ by dπ‘ is simply one. This makes the second integral much simpler. This means dπ£ by dπ‘ must be cos of two π‘. Then, the antiderivative of cos two π‘ gives us π£. Itβs a half sin of two π‘.

Now, of course, weβre working with π‘ as our parameter instead of π₯. But otherwise, the substitutions are the same. We have π’ times π£. Thatβs π‘ times a half sin two π‘, which we will shortly evaluate between zero and π by four minus the definite integral between those limits of a half sin two π‘ times dπ’ by dπ‘, which is one. We simplify this a little. And we see that weβre going to need to work out the definite integral between zero and π by four of a half of sin two π‘ with respect to π‘. Well, the antiderivative of a half sin two π‘ is negative a quarter cos two π‘. And if we so wish, we can combine these terms and then evaluate them between the limits of zero and π by four. When we substitute π by four, we get π over eight plus zero. And when we substitute zero, we get zero plus a quarter. So that gives us π by eight minus a quarter.

Letβs clear some more space and work out that final definite integral. This time, weβre going to use integration by substitution. Weβre going to let π’ be equal to sin two π‘. We know that dπ’ by dπ‘, the derivative of sin two π‘ with respect to π‘, is two cos two π‘. And whilst dπ’ by dπ‘ isnβt a fraction, we can equivalently rewrite this as a half dπ’ equals cos of two π‘ dπ‘. We then need to change our limit. So we use our substitution to do so. Our lower limit is when π‘ is equal to zero. And when π‘ is equal to zero, π’ is equal to sin of two times zero, which is zero. Then, when π‘ is equal to π by four, π’ is equal to sin of two times π by four, which is one.

We then replace sin two π‘ with π’. We can replace cos of two π‘ with a half dπ’. And of course, we must make sure we replace our limit. We might choose to take the constant factor of a half outside our integral and focus on integrating π’ squared itself. That gives us π’ cubed over three. By substituting our limits, we get a half times one cubed over three minus zero cubed over three, which is simply equal to one-sixth. Now that weβve evaluated the integrals of each of our component functions, weβre ready to return this into vector form. Itβs root two minus one π’ plus π over eight minus a quarter π£ plus one-sixth π€.