Intergration questions Watch

dj1015
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Hello

I was hoping someone could help me with some questions on intergration I have to do. I rather not be handed the solutions, but if anyone could point in me in the right direction it I would appreaciaite that.

I have no idea where to start with question 1, so that where I need the help the most. And I guess questions 2 and 3 would end up looking like a inverse trig functions, but I am not sure on how to get there.


 





1) \int  sin (\pi/4 - 2x) ( limits  are  0  and  \pi/2 )



2) \int t / \sqrt 4 - t^2 dt



3) \int 1/ \sqrt 9 - x^2


thankyou in advanced.

edit: does 1) need to be done in the form of sin A - B = Sin A CosB + Cos A Sin B and then intergrate it?
Last edited by dj1015; 6 years ago
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TRLMaths
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(Original post by Davethedavedave)
Hello

I was hoping someone could help me with some questions on intergration I have to do. I rather not be handed the solutions, but if anyone could point in me in the right direction it I would appreaciaite that.

I have no idea where to start with question 1, so that where I need the help the most. And I guess questions 2 and 3 would end up looking like a inverse trig functions, but I am not sure on how to get there.


 





1) \int  sin (\pi/4 - 2x) ( limits  are  0  and  \pi/2 )



2) \int t / \sqrt 4 - t^2 dt



3) \int 1/ \sqrt 9 - x^2


thankyou in advanced.

edit: does 1) need to be done in the form of sin A - B = Sin A CosB + Cos A Sin B and then intergrate it?
I think from what you've written you're correct in saying to use the double angle formula for question 1. For question two and three, you should get the same function. You're correct in saying it is an ivnerse trig - use your formula book if you can't remember. sin^-1?
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dj1015
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(Original post by TRLMaths)
I think from what you've written you're correct in saying to use the double angle formula for question 1. For question two and three, you should get the same function. You're correct in saying it is an ivnerse trig - use your formula book if you can't remember. sin^-1?
so you can intergrate the double angle forumla?

also in question 2) does it make any difference if 't' is in the top line and not 1 when trying to reach some inverse trig function?
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sinnhy
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For Q1. cant you just use \displaystyle\int sinx dx = -cosx + c and you will just get  (1/2) cos(\pi/4 -2x)
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dj1015
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(Original post by sinnhy)
For Q1. cant you just use \displaystyle\int sinx dx = -cosx + c and you will just get  (1/2) cos(\pi/4 -2x)
should i intergrate 1) using the chain rule?
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ztibor
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(Original post by Davethedavedave)
Hello

I was hoping someone could help me with some questions on intergration I have to do. I rather not be handed the solutions, but if anyone could point in me in the right direction it I would appreaciaite that.

I have no idea where to start with question 1, so that where I need the help the most. And I guess questions 2 and 3 would end up looking like a inverse trig functions, but I am not sure on how to get there.


 





1) \int  sin (\pi/4 - 2x) ( limits  are  0  and  \pi/2 )



2) \int t / \sqrt 4 - t^2 dt



3) \int 1/ \sqrt 9 - x^2


thankyou in advanced.

edit: does 1) need to be done in the form of sin A - B = Sin A CosB + Cos A Sin B and then intergrate it?
1)
It is a simple composite trigomometric function.
You can solve it without substitution with rules following
\displaystyle \int f(ax+b) dx=\frac{F(ax+b)}{a}+C
where F is the primitive function of f f.e. for sint is -cost
Your inner function is linear and is -2x+pi/4

2)
I think it
\int \frac {t}{\sqrt{4-t^2}} dx
This is a sample for following type of integrals
\int f'(x) \cdot f^n(x) dx =\frac{f^{n+1}(x)}{n+1}+C
In your example
 \displaystyle \int \frac {t}{\sqrt{4-t^2}} dx=-\frac{1}{2} \cdot \int (-2t)\cdot (4-t^2)^{-\frac{1}{2}} dx
and use the rule
f^n(x) is the power function of f, and is not the nth derivative which is f^{(n)}(x)
3)
There is no variable of x in the numerator , but it can be formed to the following base integral
\displaystyle  \int \frac{1}{\sqrt{1-t^2}} dt = arcsin t +C
In your example
\displaystyle \int \frac{1}{\sqrt{9-x^2}} dx=\frac{1}{3}\cdot \int \frac{1}{\sqrt{1-\left (\frac{x}{3}\right )^2}} dx =
\displaystyle = -\frac {3}{9} \int -\frac{1}{3} \cdot \frac {1}{\sqrt{1-\left (\frac{x}{3}\right )^2}} dx
Use the base integral and that
\int a\cdot f(ax) dx =F(ax)+C
whichis a special cace for that I wrote at the point 1).
Last edited by ztibor; 6 years ago
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sinnhy
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(Original post by Davethedavedave)
should i intergrate 1) using the chain rule?
yeah
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