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D) Given x = Acos(St), find:dx dtzGiven x = Acos(Vk/m 4) , find:dxdx dtz...

Question

D) Given x = Acos(St), find:dx dtzGiven x = Acos(Vk/m 4) , find:dxdx dtz

D) Given x = Acos(St), find: dx dtz Given x = Acos(Vk/m 4) , find: dx dx dtz



Answers

Integrate $\int x \sqrt{4-x} d x$ (a) by parts, letting $d v=\sqrt{4-x} d x$. (b) by substitution, letting $u=4-x$.

Who is equal to X d. U is equal to d X D. V is equal to the square root of four minus x. The X on V is equal to negative to third times four minus X to the power of three over two. So solving this out we get negative two x over three times four minus X to the power of three over two plus the integral of to third 44 minus acts to the power of three over to D X. Integrating this we end factoring out the negative to over 15 4 minus extra party over to we end up with negative to over 15 four minus X to the power of three over two times three x plus three x plus eight plus seat. So the second way we're gonna do this is we're gonna let you equal for minus acts so that d'you is than equal to negative t X. So if solving this we have the integral of you to the power of three over to minus four yoon to the power off 1/2 you integrating these two parts separately on, then factoring out to over 15 years with the power of three over two. We end up with two over 15 you to the power of three over two times three You minus 20 plus seat back substituting you on, then simplifying little bit. We end up with negative to over 15 four, minus acts to the power of three over two times three x plus eight plus C.

Suppose you evaluate the integral of X coast and of four X dx. Using integration by parts with U equal to X and DV which is co sign of four X dx. Now integration by Parts states that the integral of you D V. This is equal to U V minus the integral of V D you. So if you is X then we say d'you is dx. And if dvs clothes and a four accident integrating both sides, you get fee which is equal to Sign of four x Over four. And so the integral of X co sign of four X dx. This is equal to U V minus the integral of VD you or that's just X times Sign of four. X over four minus the integral of Sign of four x over four. The X which when simplified we have X sign of four x over four -1/4 integral of sine of four X dx. And then from here we get X sign of four x over four minus 1/4 times the integral of sine, which is co sign of four X, negative All over four. And then plus C. Now, if we factor out 1/16, we have four x sign of forex Plus co sign of four x plus C.

The given function is integral of x rays to the power minus for dx. Now we have to find the indefinite integral. Here it is of the form integral of x rays to the power N. D. X, which can be written as one divide by n plus one, Multiplied by experience to the power and plus one plus. See now apply this rule. So our function integral of x rays to the power minus four D. X. Can be written, it's one divide by minus four plus one, multiplied by x rays to the power minus four plus one plus C. After simplify, we get minus one divided by three x rays to the power three plus C. And this is our final answer.

In discussion we are required to find the value of integration X under road four plus X dx. With the help of integration by parts method and substitution method. So let's see how this whole discussion. Consider you with the calls to X. And the vehicles to under route four plus X dx. Therefore by the differentiation of you we can right do you with the calls to dx and by the integration of tv we can right vehicles to integration of under four plus X dx. So this will be calls to four Plus X to the power one by two plus one divided by one by two plus one into one. So the cell vehicles to two by three X-plus 4 to the power three by two. So now we have the values of U V D u n D V. Therefore we can apply the formula of integration by parts which can be written as integration. You're the vehicles to ub minus integration video. North of latitude all the values. So we get integration X under route four plus X dx recalls to you want to be that means two by three X into X-plus 4 to the power three x 2- integration video that means to buy three and two X plus four to the power three x 2 DX. And you know let's integrate this function so we get two by three X into X-plus 4 to the power three x 2 and the integration of X plus four to the power three but it will be called to X-plus 4 to the power three x 2 Plus one divided by three x 2 plus one plus a constant of integration. So when we further solve this we get two x 3 x X-plus 4 to the power three x 2 four by 15 in two X plus four to the power Bye Bye two plus constant of integration. From these two terms we can take out X-plus 4 to the power three x 2 as a common. So here we will have Who works up on three minus four x upon 15 minus 16 upon 15 plus a constant of integration. And when we further to all this we get X-plus 4 to the power three by two into six By 15 x 16 by 15 plus constant of integration. And from these two terms we can take out To buy 15 as a common. So finally we will have to buy 15 into X-plus 4 to the power three x 2 into three. X -8 plus a constant of integration. So this is the final answer for this integration. And now let's evaluate the same integration with the help of substitution method. No. Yeah. In order to solve the same integration with the help of substitution method we can consider you with the calls to four plus X. Therefore from here we will have extra pickles too, U -4 and by the differentiation of X We can right do you have physicals too. Do you north substitute all these values in the above integration. So we have integration X. Under route four plus XDX is the calls to integration U -4. Into under route you do you? No Multiply under you with the U -4. So we get integration U. To the power three x 2 d. U minus four. Ingration. YouTube empowerment by two. Do you? And now let's integrate these functions. Yeah so we get U. to the power three but it will be called to U to the power three by two plus one divided by three by two plus one. And this will be close to to buy five U. To the power five by two minus four. Multiplied by To buy three U. to the power three by two plus a constant of integration. And this will be called to to buy five U. To the power five by two -8 x three U. to the power three by two plus constant of integration. And from here we can take out U. To the power three x 2. Other common. So we will have To buy five U minus eight by three plus constant of integration. And now from these two terms let's take out to buy 15 other common. So we will have to buy 15 U. To the power three x 2 into three U minus 20 Plus constant of integration. And now substitute u equals two X plus four. So finally we get integration X on the road four plus X. Ds recalls to to buy 15 X-plus 4 to the power three by two into three x -8 plus constant of integration. So this is the final answer for this problem. I hope you understand the solution. Thank you.


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