Evaluate the integrals in Exercises 33–52.
∫ from -π/4 t...

Question

Evaluate the integrals in Exercises 33–52.

∫ from -π/4 to π/4 of 6 tan⁴(x) dx

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the value of the integral evaluated from negative pi divided by 3 to positive pi divided by 3. Of 4 multiplied by tangent to the power of 4 of x dx. OK. So it appears for this particular prong, we're asked to ultimately find what the value of this specific integral is that is provided to us by the prom itself. So now that we know that we're ultimately trying to solve for this integral. What we need to do first is we need to recall and state what the even function property formula is, which, as we should recall. If F of X is equal to FX, that will mean that then, this is very important, that then the integral evaluated from negative A to A of FX. DX will be equal to 2 multiplied by the integral evaluated from 0 to a of fx dx. So With that in mind, we need to recall and note based on our prior knowledge that tangent of X is an odd function, so I put O in quotation marks to remind us, and because tangent of X is an odd function, we need to note that raising an odd function to an even power will result in an even function. So, therefore, you could safely write that tangent to the power of 4 X is an even function, so I put E in quotation marks. So once again, tangent to the power of 4 of x is even. So now, when we recall and apply the even function property formula to our integral, we will be able to write that the integral evaluated from negative pi divided by 3 to pi divided by 3 of 4 multiplied by tangent to the power of 4 of x dx. Will be equal to drum roll 2 multiplied by 4 multiplied by the integral of 02 pi divided by 3 of tangent to the power of 4 of x dx, which will be all equal to the integral. Specifically, it'll be equal to 8 multiplied by the integral evaluated from 0 to pi divided by 3 of tangent to the power of 4 x dx, whereas if we multiply 2 by 4, we get 8. So now, moving right along here, we now need to rewrite the integrand and we need to note that. Tangent to the power of 2 x is equal to c 2x minus 1. So therefore, with this knowledge in mind, we could go ahead and write that 8 multiplied by the integral evaluated from 0 to pi divided by 3. Of tangent to the power of 4. Of X D X will be equal to. 8 multiplied by the integral evaluated from 0 to pi divided by 3 of parentheses of tangent squared of x to the power of 2 d x. And this will be equal to 8 multiplied by the integral evaluated from 0 to pi divided by 3 of c square of x minus 1 in parentheses to the power of 2. DX Which will be equal to 8 multiplied by the integral evaluated from 0 to pi divided by 3 of parentheses of c to the power of 4 x minus 2 multiplied by c 2 x plus 1. D x, which will be all equal to 8 multiplied by parentheses of the integral evaluated from 0 toi divided by 3 of c cant to the power of 4 x d x minus 2 multiplied by the integral evaluated from 02 pi divided by 3. Of C cant square of XDX plus the integral evaluated from 0 to pi divided by 3 of dx, and do not forget your parentheses. So now, at this stage, we need to integrate each term, and for the first term, we will need to rewrite the integrand, and when we do that, we'll be able to write that the integral evaluated from 0 to pi divided by 3 of c can to the power of 4 x dx will be equal to the integral. Evaluated from 0 to pi divided by 3 of c cant squared. Of x multiplied by c can square of x, dx. And this will be equal to the integral evaluated from 0 to pi divided by 3 parentheses of tangent 2 x + 1 multiplied by c2 xDX. So now at this point. We need to recall and use the substitution method and let us state for this particular problem that you will be equal to tangent of X, so U is equal to tangent of x. Then that will mean that DU by DX will be equal to C2 X, which will mean that DU will be equal to C2XDX. So now, our next step is we need to rewrite our limits. So, as we should note, X is equal to 0, so that will mean that if we plug in a value of 0 in for X, that will mean that U is equal to tangent of 0, and based on trigonometry, tangent of 0 is equal to 0. And X is equal to pi divided by 3 will mean that you will be equal to tangent of pi divided by 3, which will be equal to the square root of 3. Fantastic. So now we need to substitute these values back into our integral, and then we will need to recall and apply the power rule. And the power rule states that the integral of u to the power of nduu is equal to u to the power of n + 1 divided by n + 1. Plus C, where C denotes a constant variable. Where, most importantly, the variable n cannot equal -1. So with that in mind. We'll be able to write that the integral evaluated from 0 to pi divided by 3 of parentheses of tangent square of x + 1. Multiplied by. See can't squared. Of X D X will be equal to, so drum roll, this will be equal to dun da da da, the square root, so it's gonna be equal to the integral of 0 to the power of the square root of 3. So once again, We are applying the power rule and we're substituting our new limits into our integral. So when we do that, we will find that it's equal to the integral evaluated from 0 to the square root of 3 of parentheses of u to the power of 2 + 1DU, which is equal to parentheses of u to the power of 3 divided by 3 plus u. In parentheses evaluated from 0 to the square root of 3. Which is equal to parentheses of the square root of 3 in parentheses to the power of 3 divided by 3 plus the square root of 3 minus 03 divided by 3 plus 0. Which means that this will be all equal to when we evaluate that expression by plugging into our calculator, we will find that it's simply equal to 2 multiplied by the square root of 3. Fantastic However, we now need to solve for the second term, and when we do that, we will find that 2 multiplied by the integral evaluated from 0 to pi divided by 3 of c can't to the power of 2 x dx is equal to 2 multiplied by tangent of x, evaluated from 0 to pi divided by 3. Which is equal to 2 multiplied by a parentheses of tangent of pi divided by 3. Minus tangent of 0, which is equal to 2 multiplied by the square root of 3. And we also need to solve for the third term, which states that the integral evaluated from 0 to pi divided by 3 of DX is equal to x evaluated from 0 to pi divided by 3, which is equal to pi divided by 3 minus 0, which is simply equal to pi divided by 3. So therefore, without a doubt we could write 8 multiplied by parentheses of the integral evaluated from 0 to pi divided by 3 of c cant to the power of 4 of x dx minus 2 multiplied by the integral evaluated from 0 to pi divided by 3 of. C cant square x dx plus the integral evaluated from 0 to pi divided by 3 of d of x will be equal to, so drum roll. This will be equal to 8 multiplied by parentheses of 2 multiplied by the square root of 3 minus 2 multiplied by the square root of 3 plus pi divided by 3. And when we plug in this expression into our calculator, we will find that it is all equal to 8 pi divided by 3, and that is our correct and final answer. Hooray, we did it. So going back to look at the prom itself, we can safely say, without a doubt, that our final answer for our integral will be once again, drumroll, 8 pi, so it's gonna be 8 pi divided by 3, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the integrals in Exercises 33–52.∫ from -π/4 to π/4 of 6 tan⁴(x) dx

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Evaluate the integrals in Exercises 33–52.
∫ from -π/4 to π/4 of 6 tan⁴(x) dx