What is the integral of sec^3(x)?

What is the integral of sec^3(x)?

%I=int sec^3x dx%by Integration by Pats with:
u= secx and %dv=sec^2x

dx%
=> du=secx tanx dx and v=tanx,%=secxtanx-int sec x tan^2x

dx%by %tan^2x=sec^2x-1%%=secxtanx-int (sec^3x-secx) dx%since %int

sec^3xdx=I%,%=secxtanx-I+int sec x dx%by adding I and %int sec x

dx=ln|secx+tanx|+C_1%%=>2I=secxtanx+ln|secx+tanx|+C_1%by dividing by

2,%=>I=1/2secxtanx+1/2ln|secx+tanx|+C_1/2%Hence, %int sec^3

dx=1/2secxtanx+1/2ln|secx+tanx|+C%


I hope that this was helpful.

Another facile approach to the above query has been solved below with proper explanations:

Where Integration is defined as;

In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus; its inverse operation, differentiation, is the other.

Now on to the question; What is the integral of Sec^3 (x)?

I=∫ sec^3x dx

Solution:

First we are going to apply the Integration by parts. So I first am going to present the simple by parts formula. By putting the right factor:

Integration by Parts:

∫ udv = uv - ∫ vdu eq-1

We don’t have the integral of sec^3 (x) so we will have to break it up into two multiple factor;

i.e.

sec^3(x) = sec(x).sec^2(x)

∫sec^3(x)dx = ∫ sec(x) . sec^2(x)dx eq-2

Comparing equation (1) and (2)

u =sec (x) ; dv= sec^2(x)dx

We have to differentiate the u-factor and integral of dv-factor;

du/dx = d/dx . (sec x) ; ∫ dv= ∫sec^2(x) dx

du=sec(x) tan(x)dx ; v = tan (x)

Put in Equation (1):

∫udv = sec(x) tan(x) - ∫ tan(x) . [ sec(x) tan(x)] dx

Then multiplying equation on right hand side

∫udv = sec (x) tan (x) - ∫ tan^2 (x) . sec (x) dx

Using Trignometric identities of tan^2(x)

Reason: Because if we had direct integrate the above equation the process would have continued onwards.

1 + tan^2 (x) = sec^2 (x) —> tan^2 (x) = (sec^2 x – 1)

By putting the value if tan^2(x),

∫udv = sec (x) tan (x) - ∫ (sec^2 x – 1) sec (x) dx

R.H.S. of the integration à Multipily

∫udv = sec (x) tan (x) - ∫ sec^3(x) – sec (x) dx

Separation of integral:

∫udv = sec (x) tan (x) - ∫ sec^3 (x) dx + ∫ sec (x) dx

Now the integral of;

∫udv = ∫ sec^3 (x) dx

So,

∫sec^3 (x) dx = sec (x) tan (x) - ∫ sec^3 (x) dx + ∫ sec (x) dx

On R.H.S. ∫ sec^3 (x) dx to L.H.S. so both will add it.

2∫sec^3(x) dx = sec (x) tan (x) + ∫ sec (x) dx

2∫sec^3(x) dx = sec (x) tan (x) + ln │sec (x) + tan (x) │ + c

Dividing 2 on both sides we will get our answer:

∫ sec^3 (x) = ½ [ sec (x) tan (x) + ln │ sec (x) + (x) │ + c ] ------ Answer

Conclusion:

In the similar way if any question is in High Order Power so we can break it in a way that we get the Integral from one part while the other part gives us the derivative.

And if in case while performing the steps such an integral occurs that is being asked in the question so what we can do is we place it on the other side with the respective sign and continue with the solutions, while completing rest of the steps in order.

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