Integral of x(tan^2(x)) dx
The solution
Detail solution
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=x and let dv(x)=tan2(x).
Then du(x)=1.
To find v(x):
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Rewrite the integrand:
tan2(x)=sec2(x)−1
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Integrate term-by-term:
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∫sec2(x)dx=tan(x)
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The integral of a constant is the constant times the variable of integration:
∫(−1)dx=−x
The result is: −x+tan(x)
Now evaluate the sub-integral.
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫(−x)dx=−∫xdx
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The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
So, the result is: −2x2
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Rewrite the integrand:
tan(x)=cos(x)sin(x)
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Let u=cos(x).
Then let du=−sin(x)dx and substitute −du:
∫u1du
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The integral of a constant times a function is the constant times the integral of the function:
∫(−u1)du=−∫u1du
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The integral of u1 is log(u).
So, the result is: −log(u)
Now substitute u back in:
−log(cos(x))
The result is: −2x2−log(cos(x))
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Now simplify:
−2x2+xtan(x)+log(cos(x))
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Add the constant of integration:
−2x2+xtan(x)+log(cos(x))+constant
The answer is:
−2x2+xtan(x)+log(cos(x))+constant
The answer (Indefinite)
[src]
/
| 2
| 2 x
| x*tan (x) dx = C + -- + x*(-x + tan(x)) + log(cos(x))
| 2
/
2sin2(2x)+2cos2(2x)+4cos(2x)+2(sin2(2x)+cos2(2x)+2cos(2x)+1)log(sin2(2x)+cos2(2x)+2cos(2x)+1)−x2sin2(2x)+4xsin(2x)−x2cos2(2x)−2x2cos(2x)−x2
The graph
/ 2 \
1 log\1 + tan (1)/
- - - ---------------- + tan(1)
2 2
2sin22+2cos22+4cos2+2(sin22+cos22+2cos2+1)log(sin22+cos22+2cos2+1)−sin22+4sin2−cos22−2cos2−1−2log4
=
/ 2 \
1 log\1 + tan (1)/
- - - ---------------- + tan(1)
2 2
−2log(1+tan2(1))−21+tan(1)
Use the examples entering the upper and lower limits of integration.