How do you integrate #(tanx/secx) dx#?

Answer 1

The integral is #y = -cosx+ C#.

Start by simplifying the function using the identities #tantheta = sintheta/costheta# and #sectheta = 1/costheta#.
#tanx/secx = (sinx/cosx)/(1/cosx) = sinx#
If you know your basic integrals, #int(sinx)dx = -cosx + C#.

Hopefully this helps!

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Answer 2

To integrate ( \frac{\tan(x)}{\sec(x)} ) with respect to ( x ), you can first rewrite it in terms of sine and cosine functions. Using trigonometric identities, we have:

[ \frac{\tan(x)}{\sec(x)} = \frac{\sin(x)}{\cos(x) \cdot \frac{1}{\cos(x)}} = \sin(x) \cdot \cos(x) ]

Now, you can integrate ( \sin(x) \cdot \cos(x) ) using the substitution method. Let ( u = \sin(x) ), then ( du = \cos(x) , dx ). So the integral becomes:

[ \int \sin(x) \cdot \cos(x) , dx = \int u , du ]

Now, integrating ( u ) with respect to ( u ) gives:

[ \int u , du = \frac{u^2}{2} + C ]

Where ( C ) is the constant of integration.

Substituting back ( u = \sin(x) ), we have:

[ \frac{\sin^2(x)}{2} + C ]

So the integral of ( \frac{\tan(x)}{\sec(x)} ) with respect to ( x ) is ( \frac{\sin^2(x)}{2} + C ).

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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.

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.

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.

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.

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