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    Integrate cosx- sinx / 1+sin2x pls?
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    (Original post by Leon7308)
    Integrate cosx- sinx / 1+sin2x pls?
    I assume what you're trying to integrate here is (cos x - sin x)/ (1 + sin 2x).

    Two main trig formulas are needed here:
    cos^2(x)+sin^2(x) = 1 and
    sin (2x) = 2 cos (x) sin (x)

    Once you do these two subs, you'll have a fraction in a format where the lower bit of the fraction has something squared (X^2), and the upper bit has the differentiation of that something (X). Meaning that with some modifications, you can have the integral of ln (X) there!

    Multiply and divide by what's needed, and a few cancellations of terms will come along to leave you with one term to integrate!

    Let me know if you need more hints.
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    (Original post by candyaljamila)
    I assume what you're trying to integrate here is (cos x - sin x)/ (1 + sin 2x).

    Two main trig formulas are needed here:
    cos^2(x)+sin^2(x) = 1 and
    sin (2x) = 2 cos (x) sin (x)

    Once you do these two subs, you'll have a fraction in a format where the lower bit of the fraction has something squared (X^2), and the upper bit has the differentiation of that something (X). Meaning that with some modifications, you can have the integral of ln (X) there!

    Multiply and divide by what's needed, and a few cancellations of terms will come along to leave you with one term to integrate!

    Let me know if you need more hints.

    In other words, the denominator will be of the form (a + b)^(2) = a^(2) + 2ab + b^(2).

    Once the numerator and denominator are sorted out you can use a substitution of the form u = ax + bx which leads to a nice solvable integral.

    This trick crops up quite often in Trig differentiation and integration. Another example is tan(x) + 2lnsec(x).
    Differentiating leads to sec^2(x) + 2tan(x). Using tan^2(x) + 1 = sec^2(x), you have tan^2(x) + 2tan(x) + 1. This
    is again a disguised form of a^(2) + 2ab + b^(2) = (a + b)^(2). So you can write it as (tan(x) + 1)^(2).

    If you get stuck on an integration or proof that 'looks impossible', it may help to remember the trick!
 
 
 
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