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Find the indefinite integral: (Use C for the constant of integration:)sin3 40 cos 40 d0...

Question

Find the indefinite integral: (Use C for the constant of integration:)sin3 40 cos 40 d0

Find the indefinite integral: (Use C for the constant of integration:) sin3 40 cos 40 d0



Answers

$35-40$ Evaluate the integral.
$$\int_{0}^{\pi / 4}\left(\sec t \tan t \mathbf{i}+t \cos 2 t \mathbf{j}+\sin ^{2} 2 t \cos 2 t \mathbf{k}\right) d t$$

Okay, So to find the indefinite integral of co second squared overcoat tangent, I'm going to use U substitution. So I'm gonna let you be co tangent of t And then do you is gonna be negative Coast Seacon squared t d t. So if we now saw for coast second squared T d T that's going to be equal to negative d'you. So what we have is the integral of negative one over you, d you? So then this is gonna be so just take the anti derivative of negative one over you, which is gonna be negative natural law of you, plus some arbitrary, constant c So now if we bring back you you is co tangent t So our result is gonna be negative natural log of co tangent of T plus some arbitrary constants C and this is gonna be a resolve for this problem.

Uh as you look at this integral of 40 minus to oversee king of data. D Theta what I would do is just rewrite this. Um I want to do anything with that. 40 But one over 2nd is the same thing as co sign. But you still have to have that too in there. So I would just multiply that too in front. And the reason why this should be all you need to help you get the right answer is think about how the derivative of 43 data would give you 40. Um and this might be a type of if this is supposed to be 4th data kind of as I had, you would add one to the expo and divide by your new exponents would be Tuesday disc wired. I don't know if that was a typo but then think about how the derivative of sine would give you co sign and then that negative to just goes along for the ride. So um yeah I'm just double checking that the derivative of what I have in green matches what's up here. And then don't forget about your concept because the derivative of a constant is zero. But we don't write plus zero up here. So you could have any constantly. We could possibly think them there. And this will be a perfect answer unless I have a type of right here, which I kind of addressed earlier.

To find the integral of sine of two X times Cosine of four X. Dx. We apply the product to sum identity of Sinek and Co sign B. Note that sign of eight times goes N F B. This is just one half times sign of A plus B plus you have sign of A minus B. And so you see this identity we have integral of one half times A. Being two X and B being four X. We have sign of two X plus four X plus we have sign of two X minus four X. And then dx that's the same as one half integral. Love sign of six X plus sign of negative two X. The X now sign of negative to access just the same as negative sign of two X. And so we're writing this we have one half integral of sign of six X minus. Sign up two X. And then the X. Now integrating term by term we have one half times that's negative co sign of six X over six minus negative. Co sign F two X over two and then plus C. Simplifying this, we have negative co sign of six X over 12 plus. We have co sign of two X over four plus C is the same as Cosette F two X over four minus, go sign of six X over 12 and then plus C.

Let's use the reduction formula right away, which is given in question about 39. So we use that in our case, the value off incomes outers five. So it just becomes one of China's and one of the fight, and we have cost rates to the power forex. I am Sinus plus and minus one over on us. Five minutes one of five way have a bigger log powers reduced by two. So we have cost you X dia's. This can be re written husband over five cause trees to the past four x plain x we ever get plus four or five now costs Cube. There's a formula that cost three xs four cause Cube X minus three Cossacks, this can be used to replace cause cube access. Uh, cost three x three Cossacks. Okay, over for yeah, DX. And this can be easily integrated. Now this four gets canceled. So what are we left with? We have one of our five cause trees to the power four x cynics plus in plus 1/5. Integral off cost. Three Xs sine three x over three and three. Course access. Three cynics and then we have a constant often degrees. This is the final answer


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