How would you integrate this pls help asap

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Htx_x346
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#1
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Idk how to integrate that. What would be the fastest method?Name:  725D40D5-0609-430B-8211-01BADF2C4875.jpeg
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crashcody
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Start with the sin2x formula. Square it then think about getting it in terms of cos. Then, you can use reverse chain rule with the sin.
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Cocaiaaa
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(Original post by Htx_x346)
Idk how to integrate that. What would be the fastest method?Name:  725D40D5-0609-430B-8211-01BADF2C4875.jpeg
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Is this just A2 a level maths. If it is I’m scared for my life 😟
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Htx_x346
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(Original post by crashcody)
Start with the sin2x formula. Square it then think about getting it in terms of cos. Then, you can use reverse chain rule with the sin.



What about the sint at the end?
Cause we end up with sin^3t cos^2t I believe


(Original post by Cocaiaaa)
Is this just A2 a level maths. If it is I’m scared for my life 😟
Nah
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crashcody
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(Original post by Htx_x346)
What about the sint at the end?
Cause we end up with sin^3t cos^2t I believe



Nah
Yes, but don’t include the sin yet. Ill just give it away, so spoilers. Basically replace the sin^2 with 1-cos^2. Then you have a part with cos squared and cos ^4 etc. then multiple out with the sin and you should have two parts with a sin on the outside. Then you can use reverse chain rule.
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Driving_Mad
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(Original post by Htx_x346)
Nah
If you end up with sin^3tcos^2t, you could just re write this as sintcos^2tsin^2t, replace the sin^2t with 1 - cos^2t, and then integrate 4sintcos^2t - 4sintcos^4t
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Htx_x346
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(Original post by crashcody)
Yes, but don’t include the sin yet. Ill just give it away, so spoilers. Basically replace the sin^2 with 1-cos^2. Then you have a part with cos squared and cos ^4 etc. then multiple out with the sin and you should have two parts with a sin on the outside. Then you can use reverse chain rule.
(Original post by Driving_Mad)
If you end up with sin^3tcos^2t, you could just re write this as sintcos^2tsin^2t, replace the sin^2t with 1 - cos^2t, and then integrate 4sintcos^2t - 4sintcos^4t
Thanks
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Htx_x346
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Name:  1D8F2E59-7081-4C92-9F37-617F143C2A6A.jpeg
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Size:  29.6 KBOkay idk what ive done wrong but im gettting a different answer to wolframalpha
I got cos^3 t -cos^5 t
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Skiwi
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(Original post by Htx_x346)
Name:  1D8F2E59-7081-4C92-9F37-617F143C2A6A.jpeg
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Size:  29.6 KBOkay idk what ive done wrong but im gettting a different answer to wolframalpha
I got cos^3 t -cos^5 t
I got something close to what you got, though I had coefficients Infront of my trig and different negative signs. Remember Wolfram will solve it differently, when I put it into symbolabs it also gives a third different solution, all of them should be able to rearrange into the other.

You could have solved it by getting to a point where you had a sin^5(t), then expanding out using demoivres and integrating. The answer would still be correct, just looks very different. Which is what I believe symbolabs did, no idea what Wolfram did
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davros
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(Original post by Htx_x346)
Name:  1D8F2E59-7081-4C92-9F37-617F143C2A6A.jpeg
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Size:  29.6 KBOkay idk what ive done wrong but im gettting a different answer to wolframalpha
I got cos^3 t -cos^5 t
The hint in post #6 is the way to do this As noted, you should have coefficients in front of your trig functions!
(Original post by Skiwi)
I got something close to what you got, though I had coefficients Infront of my trig and different negative signs. Remember Wolfram will solve it differently, when I put it into symbolabs it also gives a third different solution, all of them should be able to rearrange into the other.

You could have solved it by getting to a point where you had a sin^5(t), then expanding out using demoivres and integrating. The answer would still be correct, just looks very different. Which is what I believe symbolabs did, no idea what Wolfram did
If you write \cos(2t) = 2\cos^2t - 1, then 3\cos2t - 7 = 6\cos^2t - 10, so Wolfram's answer multiplies out to \dfrac{4}{5}\cos^5t - \dfrac{4}{3}\cos^3t, which is what I get when I integrate using the hint given in post #6
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Skiwi
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(Original post by davros)
The hint in post #6 is the way to do this As noted, you should have coefficients in front of your trig functions!


If you write \cos(2t) = 2\cos^2t - 1, then 3\cos2t - 7 = 6\cos^2t - 10, so Wolfram's answer multiplies out to \dfrac{4}{5}\cos^5t - \dfrac{4}{3}\cos^3t, which is what I get when I integrate using the hint given in post #6
Yeh I reached the 4/5 cos^5.......
Is there any reason particular Wolfram gives the answer as that? Or is it just made to simplify things a certain way
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davros
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(Original post by Skiwi)
Yeh I reached the 4/5 cos^5.......
Is there any reason particular Wolfram gives the answer as that? Or is it just made to simplify things a certain way
I don't know what its internal algorithms are set to do tbh - I'm still amazed that you can plug random functions into an internet site and it will do this crazy stuff for you But there are quite often alternative forms for an integral (e.,g, logarithms instead of inverse hyperbolics, various equivalent trig identities etc) so don't panic too much if you get an alternative answer
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