# How do you use substitution to integrate #sec(v + pi/2) * tan(v + pi/2)#?

See the explanation.

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To integrate ( \sec(v + \frac{\pi}{2}) \cdot \tan(v + \frac{\pi}{2}) ) using substitution, you can let ( u = v + \frac{\pi}{2} ). Then, ( du = dv ) since the derivative of ( v + \frac{\pi}{2} ) with respect to ( v ) is 1.

Substituting ( u = v + \frac{\pi}{2} ), the integral becomes:

[ \int \sec(u) \tan(u) , du ]

Now, you can use the identity ( \sec(u) \tan(u) = \frac{d}{du}(\sec(u)) ).

So, integrating ( \sec(u) \tan(u) , du ) becomes integrating ( \frac{d}{du}(\sec(u)) , du ), which is simply ( \sec(u) ) plus a constant.

Substituting back ( u = v + \frac{\pi}{2} ), the result is:

[ \sec(v + \frac{\pi}{2}) + C ]

Where ( C ) is the constant of integration.

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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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