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Maths Question Help

I have been given a selection of maths questions to do over the period we are off but I have two which I have no idea where to even start. I would be very grateful if someone would be able to guide me through them. The questions are as follows (the second was harder to format on here so I hope you can understand it):

1. 50𝑐𝑜𝑠^2 𝜃+5𝑐𝑜𝑠𝜃=6𝑠𝑖𝑛^2 𝜃, where 𝜃 is acute.
Find the exact value of 𝑡𝑎𝑛𝜃
Fully justify your answer [7marks]

2. Prove that ∫1^9 1/(√𝑥(1+√(𝑥))) 𝑑𝑥=𝑎 ln⁡𝑎
Where 𝑎 is a positive integer to be found [6 marks]
Hi, I moved your thread to a more specific forum :h: hopefully people with more knowledge will see this better
Original post by hconn
I have been given a selection of maths questions to do over the period we are off but I have two which I have no idea where to even start. I would be very grateful if someone would be able to guide me through them. The questions are as follows (the second was harder to format on here so I hope you can understand it):

1. 50𝑐𝑜𝑠^2 𝜃+5𝑐𝑜𝑠𝜃=6𝑠𝑖𝑛^2 𝜃, where 𝜃 is acute.
Find the exact value of 𝑡𝑎𝑛𝜃
Fully justify your answer [7marks]

2. Prove that ∫1^9 1/(√𝑥(1+√(𝑥))) 𝑑𝑥=𝑎 ln⁡𝑎
Where 𝑎 is a positive integer to be found [6 marks]

1. Use cos2θ+sin2θ=1cos^2{\theta}+sin^2{\theta}=1 to get a quadratic in cos2θcos^2{\theta}
2. Not understanding what the equation is.
Reply 3
It's hard to show without being able to insert the full equation, the 9 is on the top of the function symbol, 1 on the bottom, you then have 1 over root x open bracket 1+ root x) and after that dx=alna. I don't know if that makes it any clearer. Thanks for the first bit though
Original post by hconn
It's hard to show without being able to insert the full equation, the 9 is on the top of the function symbol, 1 on the bottom, you then have 1 over root x open bracket 1+ root x) and after that dx=alna. I don't know if that makes it any clearer. Thanks for the first bit though

Can you upload a photo of the question?
Reply 5
Original post by Muttley79
Can you upload a photo of the question?

Ahh, I hadn't seen you could do that. Picture1.png
Original post by hconn
Ahh, I hadn't seen you could do that. Picture1.png

Have you done (a) now?
Reply 7
Original post by Muttley79
Have you done (a) now?

I think so, yes
Original post by hconn
I think so, yes

OK great - any thoughts on (b) - which integration techniques might you need/have you used?
Reply 9
Original post by Muttley79
OK great - any thoughts on (b) - which integration techniques might you need/have you used?

I don't have a clue to be honest, I feel like if I knew where to start I would have more of an idea where to go with it
Original post by hconn
I don't have a clue to be honest, I feel like if I knew where to start I would have more of an idea where to go with it


So which integration techniques do you know? Partial fractions? Substitution? I don;t want to suggest something you know nothing about
Reply 11
I know partial fractions, chain rule, integration with trig, integration by parts (but a bit hesitant on it), and there is one more but I can never remember the name for it
Original post by hconn
I know partial fractions, chain rule, integration with trig, integration by parts (but a bit hesitant on it), and there is one more but I can never remember the name for it

You can easily apply the substitution technique. Set u = 1+√x, du=1/2√x dx :. dx= 2√x du. Simplify and integrate.
Reply 13
Original post by ThiagoBrigido
You can easily apply the substitution technique. Set u = 1+√x, du=1/2√x dx :. dx= 2√x du. Simplify and integrate.

Thanks! I think I've got an answer for it now

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