integeral cos 2xdx/under root 1+sin2x
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Answer:sinx +cosx
Step-by-step explanation:
Cos2x / rt(1+ sin2x)
=cos²x - sin²x / rt(1 + 2sinxcosx)
=cos²x - sin²x / rt(sin²x + cos²cos x + 2sinxcosx)
=cos²x - sin²x / rt(sinx + cosx)²
=cos²x - sin²x / sinx + cosx
= cosx - sinx
Now integral( cosx - sinx) = sinx + cosx
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