How to Convert Sine and Cosine to Secant for Integration?

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techniques of inegration, please help

in the attachment there is a problem related to techniques of integration, i fully understand everything until near the end of the problem they answer says

(1/6)o - (1/12)sin2o + C = (1/6)o - (1/6)sino coso + C

and then they convert the sinocoso to sec-1

this part is really confusing...can you help me please...

thank you
 
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abot said:
In the attachment there is a problem related to techniques of integration, i fully understand everything until near the end of the problem they answer says
(1/6)o - (1/12)sin2o + C = (1/6)o - (1/6)sino coso + C
?
But... there's no attachment? :confused:
 
Do you know the "double angle formula":
sin(2\theta)= 2sin(\theta)cos(\theta)

That was what was used.

I'm not clear what you mean by "convert the sin(\theta)cos(\theta) to sec-1"
Of course, sec(\theta) is defined as \frac{1}{cos(\theta)} so if that "-1" is meant as reciprocal rather than "inverse function", it is true that
sin(\theta)cos(\theta)= \frac{sin(\theta)}{sec(\theta)}
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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