Integral of sec^4x dx
The solution
Detail solution
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Rewrite the integrand:
sec4(x)=(tan2(x)+1)sec2(x)
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There are multiple ways to do this integral.
Method #1
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Let u=tan(x).
Then let du=(tan2(x)+1)dx and substitute du:
∫(u2+1)du
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Integrate term-by-term:
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The integral of un is n+1un+1 when n=−1:
∫u2du=3u3
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The integral of a constant is the constant times the variable of integration:
∫1du=u
The result is: 3u3+u
Now substitute u back in:
3tan3(x)+tan(x)
Method #2
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Rewrite the integrand:
(tan2(x)+1)sec2(x)=tan2(x)sec2(x)+sec2(x)
-
Integrate term-by-term:
-
Let u=tan(x).
Then let du=(tan2(x)+1)dx and substitute du:
∫u2du
-
The integral of un is n+1un+1 when n=−1:
∫u2du=3u3
Now substitute u back in:
3tan3(x)
-
∫sec2(x)dx=tan(x)
The result is: 3tan3(x)+tan(x)
Method #3
-
Rewrite the integrand:
(tan2(x)+1)sec2(x)=tan2(x)sec2(x)+sec2(x)
-
Integrate term-by-term:
-
Let u=tan(x).
Then let du=(tan2(x)+1)dx and substitute du:
∫u2du
-
The integral of un is n+1un+1 when n=−1:
∫u2du=3u3
Now substitute u back in:
3tan3(x)
-
∫sec2(x)dx=tan(x)
The result is: 3tan3(x)+tan(x)
-
Add the constant of integration:
3tan3(x)+tan(x)+constant
The answer is:
3tan3(x)+tan(x)+constant
The answer (Indefinite)
[src]
/
| 3
| 4 tan (x)
| sec (x) dx = C + ------- + tan(x)
| 3
/
∫sec4(x)dx=C+3tan3(x)+tan(x)
The graph
sin(1) 2*sin(1)
--------- + --------
3 3*cos(1)
3*cos (1)
3cos(1)2sin(1)+3cos3(1)sin(1)
=
sin(1) 2*sin(1)
--------- + --------
3 3*cos(1)
3*cos (1)
3cos(1)2sin(1)+3cos3(1)sin(1)
sin(1)/(3*cos(1)^3) + 2*sin(1)/(3*cos(1))
Use the examples entering the upper and lower limits of integration.