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

Answer 1

See the explanation.

#t=1/sec(v+pi/2)#
#dt=((sec(v+pi/2))^-1)'dv=#
#dt=-1/sec^2(v+pi/2)sec(v+pi/2)tan(v+pi/2)dv#
#dt=-(tan(v+pi/2)dv)/sec(v+pi/2)=-t tan(v+pi/2)dv#
#tan(v+pi/2)dv=-dt/t#
#int sec(v+pi/2) tan(v+pi/2) dv=int 1/t(-dt/t)=#
#=-int t^-2dt=1/t+C=sec(v+pi/2)+C#
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Answer 2

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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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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