Integral of sec^2xtanx dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Let u=sec2(x).
Then let du=2tan(x)sec2(x)dx and substitute 2du:
∫41du
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The integral of a constant times a function is the constant times the integral of the function:
∫21du=2∫1du
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The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: 2u
Now substitute u back in:
2sec2(x)
Method #2
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Let u=sec(x).
Then let du=tan(x)sec(x)dx and substitute du:
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The integral of un is n+1un+1 when n=−1:
∫udu=2u2
Now substitute u back in:
2sec2(x)
Method #3
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Let u=tan(x).
Then let du=(tan2(x)+1)dx and substitute du:
-
The integral of un is n+1un+1 when n=−1:
∫udu=2u2
Now substitute u back in:
2tan2(x)
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Add the constant of integration:
2sec2(x)+constant
The answer is:
2sec2(x)+constant
The answer (Indefinite)
[src]
/
| 2
| 2 sec (x)
| sec (x)*tan(x) dx = C + -------
| 2
/
2tan2x
The graph
1 1
- - + ---------
2 2
2*cos (1)
−2sin21−21−21
=
1 1
- - + ---------
2 2
2*cos (1)
−21+2cos2(1)1
Use the examples entering the upper and lower limits of integration.