Integral of tan^3xsec^3x dx
The solution
Detail solution
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Rewrite the integrand:
tan3(x)sec3(x)=(sec2(x)−1)tan(x)sec3(x)
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There are multiple ways to do this integral.
Method #1
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Let u=sec(x).
Then let du=tan(x)sec(x)dx and substitute du:
∫(u4−u2)du
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Integrate term-by-term:
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The integral of un is n+1un+1 when n=−1:
∫u4du=5u5
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The integral of a constant times a function is the constant times the integral of the function:
∫(−u2)du=−∫u2du
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The integral of un is n+1un+1 when n=−1:
∫u2du=3u3
So, the result is: −3u3
The result is: 5u5−3u3
Now substitute u back in:
5sec5(x)−3sec3(x)
Method #2
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Rewrite the integrand:
(sec2(x)−1)tan(x)sec3(x)=tan(x)sec5(x)−tan(x)sec3(x)
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Integrate term-by-term:
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Let u=sec5(x).
Then let du=5tan(x)sec5(x)dx and substitute 5du:
∫251du
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The integral of a constant times a function is the constant times the integral of the function:
∫51du=5∫1du
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The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: 5u
Now substitute u back in:
5sec5(x)
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The integral of a constant times a function is the constant times the integral of the function:
∫(−tan(x)sec3(x))dx=−∫tan(x)sec3(x)dx
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Let u=sec3(x).
Then let du=3tan(x)sec3(x)dx and substitute 3du:
∫91du
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The integral of a constant times a function is the constant times the integral of the function:
∫31du=3∫1du
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The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: 3u
Now substitute u back in:
3sec3(x)
So, the result is: −3sec3(x)
The result is: 5sec5(x)−3sec3(x)
Method #3
-
Rewrite the integrand:
(sec2(x)−1)tan(x)sec3(x)=tan(x)sec5(x)−tan(x)sec3(x)
-
Integrate term-by-term:
-
Let u=sec5(x).
Then let du=5tan(x)sec5(x)dx and substitute 5du:
∫251du
-
The integral of a constant times a function is the constant times the integral of the function:
∫51du=5∫1du
-
The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: 5u
Now substitute u back in:
5sec5(x)
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−tan(x)sec3(x))dx=−∫tan(x)sec3(x)dx
-
Let u=sec3(x).
Then let du=3tan(x)sec3(x)dx and substitute 3du:
∫91du
-
The integral of a constant times a function is the constant times the integral of the function:
∫31du=3∫1du
-
The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: 3u
Now substitute u back in:
3sec3(x)
So, the result is: −3sec3(x)
The result is: 5sec5(x)−3sec3(x)
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Add the constant of integration:
5sec5(x)−3sec3(x)+constant
The answer is:
5sec5(x)−3sec3(x)+constant
The answer (Indefinite)
[src]
/
| 3 5
| 3 3 sec (x) sec (x)
| tan (x)*sec (x) dx = C - ------- + -------
| 3 5
/
−15cos5x5cos2x−3
The graph
2
2 3 - 5*cos (1)
-- + -------------
15 5
15*cos (1)
−3cos311+5cos511+152
=
2
2 3 - 5*cos (1)
-- + -------------
15 5
15*cos (1)
152+15cos5(1)3−5cos2(1)
Use the examples entering the upper and lower limits of integration.