Answer :
Sure! Let's start with the step-by-step solution for each part in Q4.
### Q4.1:
a) Evaluate the integral:
[tex]\[ \int \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right)^2 dx \][/tex]
First, simplify the integrand:
[tex]\[ \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right)^2 = \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right) \cdot \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right) \][/tex]
[tex]\[ = \frac{(\sqrt{x+2})^2}{(\sqrt{x})^2} = \frac{x+2}{x} = 1 + \frac{2}{x} \][/tex]
Now, rewrite the integral:
[tex]\[ \int \left( 1 + \frac{2}{x} \right) dx \][/tex]
[tex]\[ = \int 1 \, dx + \int \frac{2}{x} \, dx \][/tex]
Evaluate each term separately:
[tex]\[ \int 1 \, dx = x \][/tex]
[tex]\[ \int \frac{2}{x} \, dx = 2 \ln |x| \][/tex]
Combining the results, we get:
[tex]\[ x + 2 \ln |x| + C \][/tex]
Therefore,
[tex]\[ \int \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right)^2 dx = x + 2 \ln |x| + C \][/tex]
b) Evaluate the integral:
[tex]\[ \int \sin x (\cos x + 3) dx \][/tex]
First, distribute [tex]\(\sin x\)[/tex]:
[tex]\[ \int \sin x \cos x \, dx + \int 3 \sin x \, dx \][/tex]
Now, we will use substitution for each integral:
For [tex]\(\int \sin x \cos x \, dx\)[/tex], set [tex]\(u = \cos x\)[/tex]:
[tex]\[ \frac{du}{dx} = -\sin x \][/tex]
[tex]\[ du = -\sin x \, dx \][/tex]
[tex]\[ dx = -\frac{du}{\sin x} \][/tex]
Substitute:
[tex]\[ \int \sin x \cos x \, dx = \int -\cos x \, du = -\int u \, du = -\frac{u^2}{2} + C_1 = -\frac{\cos^2 x}{2} + C_1 \][/tex]
For [tex]\(\int 3 \sin x \, dx\)[/tex], we know:
[tex]\[ \int 3 \sin x \, dx = -3 \cos x + C_2 \][/tex]
Combining both results, we get:
[tex]\[ \int \sin x (\cos x + 3) dx = -\frac{\cos^2 x}{2} - 3 \cos x + C \][/tex]
### Q4.2:
a) Evaluate the integral:
[tex]\[ \int \left( \frac{1}{4 + x^2} \right) dx \][/tex]
This is a standard integral that can be recognized as a form of the arctangent function. Specifically:
[tex]\[ \int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \arctan\left( \frac{x}{a} \right) + C \][/tex]
For our case, [tex]\(a = 2\)[/tex]:
[tex]\[ \int \frac{1}{4 + x^2} dx = \frac{1}{2} \arctan\left( \frac{x}{2} \right) + C \][/tex]
b) Evaluate the integral:
[tex]\[ \int_0^2 \frac{(x-2)}{(x+3)(x-4)} dx \][/tex]
To evaluate this improper integral, use partial fraction decomposition. Assume:
[tex]\[ \frac{x-2}{(x+3)(x-4)} = \frac{A}{x+3} + \frac{B}{x-4} \][/tex]
Multiply both sides by [tex]\((x+3)(x-4)\)[/tex]:
[tex]\[ x - 2 = A(x-4) + B(x+3) \][/tex]
Now, solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex] by substituting convenient values of [tex]\(x\)[/tex]:
- Let [tex]\(x = 4\)[/tex]:
[tex]\[ 4 - 2 = A(4-4) + B(4+3) \][/tex]
[tex]\[ 2 = 7B \][/tex]
[tex]\[ B = \frac{2}{7} \][/tex]
- Let [tex]\(x = -3\)[/tex]:
[tex]\[ -3 - 2 = A(-3-4) + B(-3+3) \][/tex]
[tex]\[ -5 = -7A \][/tex]
[tex]\[ A = \frac{5}{7} \][/tex]
So the partial fraction decomposition is:
[tex]\[ \frac{x-2}{(x+3)(x-4)} = \frac{5/7}{x+3} + \frac{2/7}{x-4} \][/tex]
Rewrite the original integral:
[tex]\[ \int_0^2 \left( \frac{5/7}{x+3} + \frac{2/7}{x-4} \right) dx \][/tex]
[tex]\[ = \frac{5}{7} \int_0^2 \frac{1}{x+3} dx + \frac{2}{7} \int_0^2 \frac{1}{x-4} dx \][/tex]
Evaluate each integral separately:
For [tex]\(\int_0^2 \frac{1}{x+3} dx\)[/tex], set [tex]\(u = x + 3\)[/tex]:
[tex]\[ du = dx \][/tex]
[tex]\[ \int_3^5 \frac{1}{u} du = \ln |u| \Big|_3^5 \][/tex]
[tex]\[ = \ln 5 - \ln 3 = \ln \frac{5}{3} \][/tex]
For [tex]\(\int_0^2 \frac{1}{x-4} dx\)[/tex], set [tex]\(v = x - 4\)[/tex]:
[tex]\[ dv = dx \][/tex]
[tex]\[ \int_{-4}^{-2} \frac{1}{v} dv = \ln |v| \Big|_{-4}^{-2} \][/tex]
[tex]\[ = \ln (-2) - \ln (-4) = \ln \left( \frac{-2}{-4} \right) = \ln \frac{1}{2} = -\ln 2 \][/tex]
Combine the results and remember the integral bounds need to match (since one part becomes negative log):
[tex]\[ \frac{5}{7} \ln \frac{5}{3} + \frac{2}{7} (-\ln 2) \][/tex]
[tex]\[ = \frac{5}{7} \ln \frac{5}{3} - \frac{2}{7} \ln 2 \][/tex]
Therefore:
[tex]\[ \int_0^2 \frac{(x-2)}{(x+3)(x-4)} dx = \frac{5}{7} \ln \frac{5}{3} - \frac{2}{7} \ln 2 \][/tex]
### Q4.1:
a) Evaluate the integral:
[tex]\[ \int \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right)^2 dx \][/tex]
First, simplify the integrand:
[tex]\[ \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right)^2 = \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right) \cdot \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right) \][/tex]
[tex]\[ = \frac{(\sqrt{x+2})^2}{(\sqrt{x})^2} = \frac{x+2}{x} = 1 + \frac{2}{x} \][/tex]
Now, rewrite the integral:
[tex]\[ \int \left( 1 + \frac{2}{x} \right) dx \][/tex]
[tex]\[ = \int 1 \, dx + \int \frac{2}{x} \, dx \][/tex]
Evaluate each term separately:
[tex]\[ \int 1 \, dx = x \][/tex]
[tex]\[ \int \frac{2}{x} \, dx = 2 \ln |x| \][/tex]
Combining the results, we get:
[tex]\[ x + 2 \ln |x| + C \][/tex]
Therefore,
[tex]\[ \int \left(\frac{\sqrt{x+2}}{\sqrt{x}}\right)^2 dx = x + 2 \ln |x| + C \][/tex]
b) Evaluate the integral:
[tex]\[ \int \sin x (\cos x + 3) dx \][/tex]
First, distribute [tex]\(\sin x\)[/tex]:
[tex]\[ \int \sin x \cos x \, dx + \int 3 \sin x \, dx \][/tex]
Now, we will use substitution for each integral:
For [tex]\(\int \sin x \cos x \, dx\)[/tex], set [tex]\(u = \cos x\)[/tex]:
[tex]\[ \frac{du}{dx} = -\sin x \][/tex]
[tex]\[ du = -\sin x \, dx \][/tex]
[tex]\[ dx = -\frac{du}{\sin x} \][/tex]
Substitute:
[tex]\[ \int \sin x \cos x \, dx = \int -\cos x \, du = -\int u \, du = -\frac{u^2}{2} + C_1 = -\frac{\cos^2 x}{2} + C_1 \][/tex]
For [tex]\(\int 3 \sin x \, dx\)[/tex], we know:
[tex]\[ \int 3 \sin x \, dx = -3 \cos x + C_2 \][/tex]
Combining both results, we get:
[tex]\[ \int \sin x (\cos x + 3) dx = -\frac{\cos^2 x}{2} - 3 \cos x + C \][/tex]
### Q4.2:
a) Evaluate the integral:
[tex]\[ \int \left( \frac{1}{4 + x^2} \right) dx \][/tex]
This is a standard integral that can be recognized as a form of the arctangent function. Specifically:
[tex]\[ \int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \arctan\left( \frac{x}{a} \right) + C \][/tex]
For our case, [tex]\(a = 2\)[/tex]:
[tex]\[ \int \frac{1}{4 + x^2} dx = \frac{1}{2} \arctan\left( \frac{x}{2} \right) + C \][/tex]
b) Evaluate the integral:
[tex]\[ \int_0^2 \frac{(x-2)}{(x+3)(x-4)} dx \][/tex]
To evaluate this improper integral, use partial fraction decomposition. Assume:
[tex]\[ \frac{x-2}{(x+3)(x-4)} = \frac{A}{x+3} + \frac{B}{x-4} \][/tex]
Multiply both sides by [tex]\((x+3)(x-4)\)[/tex]:
[tex]\[ x - 2 = A(x-4) + B(x+3) \][/tex]
Now, solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex] by substituting convenient values of [tex]\(x\)[/tex]:
- Let [tex]\(x = 4\)[/tex]:
[tex]\[ 4 - 2 = A(4-4) + B(4+3) \][/tex]
[tex]\[ 2 = 7B \][/tex]
[tex]\[ B = \frac{2}{7} \][/tex]
- Let [tex]\(x = -3\)[/tex]:
[tex]\[ -3 - 2 = A(-3-4) + B(-3+3) \][/tex]
[tex]\[ -5 = -7A \][/tex]
[tex]\[ A = \frac{5}{7} \][/tex]
So the partial fraction decomposition is:
[tex]\[ \frac{x-2}{(x+3)(x-4)} = \frac{5/7}{x+3} + \frac{2/7}{x-4} \][/tex]
Rewrite the original integral:
[tex]\[ \int_0^2 \left( \frac{5/7}{x+3} + \frac{2/7}{x-4} \right) dx \][/tex]
[tex]\[ = \frac{5}{7} \int_0^2 \frac{1}{x+3} dx + \frac{2}{7} \int_0^2 \frac{1}{x-4} dx \][/tex]
Evaluate each integral separately:
For [tex]\(\int_0^2 \frac{1}{x+3} dx\)[/tex], set [tex]\(u = x + 3\)[/tex]:
[tex]\[ du = dx \][/tex]
[tex]\[ \int_3^5 \frac{1}{u} du = \ln |u| \Big|_3^5 \][/tex]
[tex]\[ = \ln 5 - \ln 3 = \ln \frac{5}{3} \][/tex]
For [tex]\(\int_0^2 \frac{1}{x-4} dx\)[/tex], set [tex]\(v = x - 4\)[/tex]:
[tex]\[ dv = dx \][/tex]
[tex]\[ \int_{-4}^{-2} \frac{1}{v} dv = \ln |v| \Big|_{-4}^{-2} \][/tex]
[tex]\[ = \ln (-2) - \ln (-4) = \ln \left( \frac{-2}{-4} \right) = \ln \frac{1}{2} = -\ln 2 \][/tex]
Combine the results and remember the integral bounds need to match (since one part becomes negative log):
[tex]\[ \frac{5}{7} \ln \frac{5}{3} + \frac{2}{7} (-\ln 2) \][/tex]
[tex]\[ = \frac{5}{7} \ln \frac{5}{3} - \frac{2}{7} \ln 2 \][/tex]
Therefore:
[tex]\[ \int_0^2 \frac{(x-2)}{(x+3)(x-4)} dx = \frac{5}{7} \ln \frac{5}{3} - \frac{2}{7} \ln 2 \][/tex]
