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How do you integrate #(1-tan2x)/(sec2x)dx#?

Answer 1

Scarlett Allison

You get: #sin(2x)/2+cos(2x)/2+c#

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Answer 2

Elijah Abram

To integrate (\frac{{1 - \tan^2 x}}{{\sec^2 x}} , dx), first simplify the integrand:

[ \frac{{1 - \tan^2 x}}{{\sec^2 x}} = \frac{{1 - \frac{{\sin^2 x}}{{\cos^2 x}}}}{{\frac{1}{{\cos^2 x}}}} = \frac{{\cos^2 x - \sin^2 x}}{{1}} = \cos(2x) ]

Now integrate (\cos(2x)) with respect to (x):

[ \int \cos(2x) , dx = \frac{1}{2} \sin(2x) + C ]

So the integral of (\frac{{1 - \tan^2 x}}{{\sec^2 x}} , dx) is (\frac{1}{2} \sin(2x) + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.