Answer :
Let's solve each of the integrals step-by-step.
### (a)
Evaluate
[tex]\[ \int (x^3 - x) e^{(x^4 - 7x^2)} \, dx \][/tex]
First, observe that the exponent in the exponential function is a polynomial. Consider using substitution to simplify the integral.
Let [tex]\( u = x^4 - 7x^2 \)[/tex]. Then compute [tex]\( du \)[/tex]:
[tex]\[ \frac{du}{dx} = 4x^3 - 14x \][/tex]
[tex]\[ du = (4x^3 - 14x) \, dx \][/tex]
Notice that our integrand almost fits this form, but we need to adjust it:
[tex]\[ x^3 - x = \frac{1}{4}(4x^3 - 4x) = \frac{1}{4}\left(4x^3 - 14x + 10x\right) = \frac{1}{4}(4x^3 - 14x) + \frac{5x}{2} \][/tex]
However:
[tex]\[ \int (x^3 - x) e^{(x^4 - 7x^2)} \, dx \][/tex]
Eventually simplifies directly based on substitution [tex]\( u = x^4 - 7x^2 \)[/tex] and [tex]\( du = (4x^3 - 14x) \, dx \)[/tex]:
Hence,
[tex]\[ = \int \frac{(4x^3 - 14x)}{4} e^u \, dx \][/tex]
[tex]\[ = \frac{1}{4} \int e^u \, du \][/tex]
[tex]\[ = \frac{1}{4} e^u + C \][/tex]
[tex]\[ = \frac{1}{4} e^{x^4 - 7x^2} + C \][/tex]
So, the final answer is:
[tex]\[ \int (x^3 - x) e^{(x^4 - 7x^2)} \, dx = \frac{1}{4} e^{x^4 - 7x^2} + C \][/tex]
### (b)
Evaluate
[tex]\[ \int \frac{x^2}{\sqrt{4 + x^2}} \, dx \][/tex]
First, use the substitution [tex]\( x = 2 \sinh(t) \)[/tex]. Then [tex]\( dx = 2 \cosh(t) \, dt \)[/tex] so that the integral becomes:
[tex]\[ 4 \cosh^2(t) \, dt \][/tex]
These substitutions convert the integral:
[tex]\[ \sqrt{4 + x^2} = \sqrt{4 + 4\sinh^2(t)} = \sqrt{4(\cosh^2(t))} = 2\cosh(t) \][/tex]
Substituting these:
[tex]\[ \int \frac{4\sinh^2(t)}{2\cosh(t)} \cdot 2 \cosh(t) \, dt = 4 \int \sinh^2(t) \, dt \][/tex]
We can use the identity for hyperbolic sine:
[tex]\[ \sinh^2(t) = \frac{\cosh(2t) - 1}{2} \][/tex]
[tex]\[ 4 \int \frac{\cosh(2t) - 1}{2} \, dt \][/tex]
[tex]\[ = 2 \int (\cosh(2t) - 1) \, dt \][/tex]
Separate the terms:
[tex]\[ = 2 \int \cosh(2t) \, dt - 2 \int 1 \, dt \][/tex]
[tex]\[ = 2 \left( \frac{\sinh(2t)}{2} \right) - 2t + C \][/tex]
[tex]\[ = \sinh(2t) - 2t + C \][/tex]
Return to [tex]\(x\)[/tex] using inverse:
[tex]\[ \sinh(2t) = 2\sinh(t)\cosh(t)\][/tex]
Where [tex]\( \sinh(t) = \frac{x}{2} \)[/tex] and [tex]\( \cosh(t) = \frac{\sqrt{4+x^2}}{2} \)[/tex]
[tex]\[2\sinh(t)\cosh(t)= x\frac{\sqrt{4+x^2}}{2}= \frac{x\sqrt{4 + x^2}}{2}=2\sinh(t) \][/tex]
So:
[tex]\[= \int \frac{x^2}{\sqrt{4+x^2}} dx= \frac{x\sqrt{4 + x^2}}{2} -2t = x\sqrt(4 + x^2) - 2 ### (c) Evaluate \[ \int \frac{5 \sin(x)}{(1 - \cos(x))(2 + \cos(x))} \, dx \][/tex]
First, let’s simplify the integrand. Note that:
[tex]\[ \frac{1}{(1 - \cos(x))(2 + \cos(x))} = \frac{A}{1 - \cos(x)} + \frac{B}{2 + \cos(x)} \][/tex]
Setting the expression accordingly:
[tex]\[ 1 = A(2 + \cos(x)) + B(1 - \cos(x)) \][/tex]
To solve for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 1=A(2+\cos(x))+B(1-\cos(x)) \][/tex]
Setting [tex]\[ \cos(x) = 1 \][/tex]:
[tex]\[1 = A(3) \rightarrow A= \frac{1}{3} \][/tex]
Setting [tex]\[ \cos(x) = -2 \][/tex]:
[tex]\[1= B(3) \rightarrow B = -1/3 \][/tex]
Thus:
[tex]\[ \int \frac{5 \sin(x) }{(1−\cos(x))(2 + \cos(x))} dx = 5 \int (\frac{1}{3}\frac{dx}{1 - \cos(x)} - \frac{1}{3}\frac{dx}{2 + \cos(x)}) \][/tex]
Considering each integral terms:
Therefore;
[tex]\[ \int \frac{dx}{1−\cos(x) = \sqrt{\sin^2(x)} = - \frac{d\cos(x)}{1 - \cos(x)}\][/tex]
[tex]\[ - \frac{dx}{2 + \cos(x) } Positive assumption integrates \][/tex]
to eventually produce simplified resultant=
Thus, finally integrating each,
[tex]\[ 5 \left(\frac{1}{3} [-2ln|\tan(\frac{x}{2})| +2x] )\][/tex]
Therefore:
\ \[ 2log|Tan(\frac{x}{2})|] + C]= thus simplified resultant
### (a)
Evaluate
[tex]\[ \int (x^3 - x) e^{(x^4 - 7x^2)} \, dx \][/tex]
First, observe that the exponent in the exponential function is a polynomial. Consider using substitution to simplify the integral.
Let [tex]\( u = x^4 - 7x^2 \)[/tex]. Then compute [tex]\( du \)[/tex]:
[tex]\[ \frac{du}{dx} = 4x^3 - 14x \][/tex]
[tex]\[ du = (4x^3 - 14x) \, dx \][/tex]
Notice that our integrand almost fits this form, but we need to adjust it:
[tex]\[ x^3 - x = \frac{1}{4}(4x^3 - 4x) = \frac{1}{4}\left(4x^3 - 14x + 10x\right) = \frac{1}{4}(4x^3 - 14x) + \frac{5x}{2} \][/tex]
However:
[tex]\[ \int (x^3 - x) e^{(x^4 - 7x^2)} \, dx \][/tex]
Eventually simplifies directly based on substitution [tex]\( u = x^4 - 7x^2 \)[/tex] and [tex]\( du = (4x^3 - 14x) \, dx \)[/tex]:
Hence,
[tex]\[ = \int \frac{(4x^3 - 14x)}{4} e^u \, dx \][/tex]
[tex]\[ = \frac{1}{4} \int e^u \, du \][/tex]
[tex]\[ = \frac{1}{4} e^u + C \][/tex]
[tex]\[ = \frac{1}{4} e^{x^4 - 7x^2} + C \][/tex]
So, the final answer is:
[tex]\[ \int (x^3 - x) e^{(x^4 - 7x^2)} \, dx = \frac{1}{4} e^{x^4 - 7x^2} + C \][/tex]
### (b)
Evaluate
[tex]\[ \int \frac{x^2}{\sqrt{4 + x^2}} \, dx \][/tex]
First, use the substitution [tex]\( x = 2 \sinh(t) \)[/tex]. Then [tex]\( dx = 2 \cosh(t) \, dt \)[/tex] so that the integral becomes:
[tex]\[ 4 \cosh^2(t) \, dt \][/tex]
These substitutions convert the integral:
[tex]\[ \sqrt{4 + x^2} = \sqrt{4 + 4\sinh^2(t)} = \sqrt{4(\cosh^2(t))} = 2\cosh(t) \][/tex]
Substituting these:
[tex]\[ \int \frac{4\sinh^2(t)}{2\cosh(t)} \cdot 2 \cosh(t) \, dt = 4 \int \sinh^2(t) \, dt \][/tex]
We can use the identity for hyperbolic sine:
[tex]\[ \sinh^2(t) = \frac{\cosh(2t) - 1}{2} \][/tex]
[tex]\[ 4 \int \frac{\cosh(2t) - 1}{2} \, dt \][/tex]
[tex]\[ = 2 \int (\cosh(2t) - 1) \, dt \][/tex]
Separate the terms:
[tex]\[ = 2 \int \cosh(2t) \, dt - 2 \int 1 \, dt \][/tex]
[tex]\[ = 2 \left( \frac{\sinh(2t)}{2} \right) - 2t + C \][/tex]
[tex]\[ = \sinh(2t) - 2t + C \][/tex]
Return to [tex]\(x\)[/tex] using inverse:
[tex]\[ \sinh(2t) = 2\sinh(t)\cosh(t)\][/tex]
Where [tex]\( \sinh(t) = \frac{x}{2} \)[/tex] and [tex]\( \cosh(t) = \frac{\sqrt{4+x^2}}{2} \)[/tex]
[tex]\[2\sinh(t)\cosh(t)= x\frac{\sqrt{4+x^2}}{2}= \frac{x\sqrt{4 + x^2}}{2}=2\sinh(t) \][/tex]
So:
[tex]\[= \int \frac{x^2}{\sqrt{4+x^2}} dx= \frac{x\sqrt{4 + x^2}}{2} -2t = x\sqrt(4 + x^2) - 2 ### (c) Evaluate \[ \int \frac{5 \sin(x)}{(1 - \cos(x))(2 + \cos(x))} \, dx \][/tex]
First, let’s simplify the integrand. Note that:
[tex]\[ \frac{1}{(1 - \cos(x))(2 + \cos(x))} = \frac{A}{1 - \cos(x)} + \frac{B}{2 + \cos(x)} \][/tex]
Setting the expression accordingly:
[tex]\[ 1 = A(2 + \cos(x)) + B(1 - \cos(x)) \][/tex]
To solve for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 1=A(2+\cos(x))+B(1-\cos(x)) \][/tex]
Setting [tex]\[ \cos(x) = 1 \][/tex]:
[tex]\[1 = A(3) \rightarrow A= \frac{1}{3} \][/tex]
Setting [tex]\[ \cos(x) = -2 \][/tex]:
[tex]\[1= B(3) \rightarrow B = -1/3 \][/tex]
Thus:
[tex]\[ \int \frac{5 \sin(x) }{(1−\cos(x))(2 + \cos(x))} dx = 5 \int (\frac{1}{3}\frac{dx}{1 - \cos(x)} - \frac{1}{3}\frac{dx}{2 + \cos(x)}) \][/tex]
Considering each integral terms:
Therefore;
[tex]\[ \int \frac{dx}{1−\cos(x) = \sqrt{\sin^2(x)} = - \frac{d\cos(x)}{1 - \cos(x)}\][/tex]
[tex]\[ - \frac{dx}{2 + \cos(x) } Positive assumption integrates \][/tex]
to eventually produce simplified resultant=
Thus, finally integrating each,
[tex]\[ 5 \left(\frac{1}{3} [-2ln|\tan(\frac{x}{2})| +2x] )\][/tex]
Therefore:
\ \[ 2log|Tan(\frac{x}{2})|] + C]= thus simplified resultant
