Answer to Question #167370 in Calculus for csdf

Question

1) Antiderivative of -4csc(x)cot(x)dx :Show all work

2) Antiderivative of -1cot^2(x)dx :Show all work

3) Antiderivative of ((6/x^3)+cube root of x^2)dx :Show all work

4) Write the Antiderivative and definite integral (leave in terms of e if applicable) --> the integral from -1 to 3 (10e^x-4x)dx

5) Write the Antiderivative and definite integral(leave in radical form if applicable)--> the integral from (pi/2) to (5pi/6) (-3cos(x)-4sin(x))dx

6) Find the Antiderivative of sec(x) tan(x)dx- if you found an antiderivative for this explain why the result for the definite integral using the same function is undefined??

7) Given the function h(x)= (square root of x)*ln(sin(x))--> Use a calculator to evaluate each of the following and round to the nearest thousandth (3 decimal places).

**Given--> h(x)= (square root of x)*ln(sin(x))

a) The integral from 0.5 to 3 h(x)dx

b)The integral from 3 to 2 h(x)dx

c)The integral from 7 to 9 |h(x)| dx

Answer 1

1) F'(x)=-4csc(x)cot(x)

$F(x)=-4\int\csc(x)\cot(x)dx=-4\int\frac{\cos x}{\sin^2x}dx=-4\int\frac{d\sin x}{\sin^2x}=\frac{4}{\sin x}+c$

2) F'(x)=-1cot^2(x)

$F(x)=-\int \cot^2(x)dx=-\int\frac{\cos^2 x}{\sin^2x}dx=\int\frac{1-\cos^2 x}{\sin^2x}dx-\int\frac{1}{\sin^2x} dx=\int dx +\int d(\cot x)=\cot x+x+c$

3) $F'(x)=\frac{6}{x^3}+\sqrt[3]{x^2}$

$F(x)=\int (\frac{6}{x^3}+\sqrt[3]{x^2})dx=-\frac{3}{x^2}+\frac{3}{5}x^{5/3}+c$

4) $F'(x)=10e^x-4x$

$F(x)=\int(10e^x-4x)dx=10e^x-2x^2+c$

$\int\limits_{-1}^3 (10e^x-4x)dx=F(3)-F(-1)=10e^{3}-2\cdot 3^2-(10e^{-1}-2)=10(e^3-e^{-1})-16$

5) $F'(x)=-3\cos x-\sin x$

$F(x)=\int(-3\cos x-4\sin x)dx=-3\sin x+4\cos x+c$

$\int\limits_{\pi/2}^{5\pi/6}(-3\cos x-4\sin x)dx=F(\frac{5\pi}{6})-F(\frac{\pi}{2})=(-3\sin \frac{5\pi}{6}+4\cos \frac{5\pi}{6})-(-3\sin \frac{\pi}{2}+4\cos \frac{\pi}{2})=-3/2+4\sqrt3/2+3-0=2\sqrt3+3/2$

6) F'(x)=sec(x) tan(x)

$F(x)=\int \sec x \tan x dx=\int\frac{\sin x}{\cos^2x}dx=-\int\frac{1}{\cos^2x}d\cos x=\frac{1}{\cos x}+c$

The result for the definite integral over the segment [a,b] is correctly defind by the formula F(b)-F(a), only if this segment does not contain a point x, such that cos(x)=0 (that is, $\pi/2+\pi n)$. If not, then the function F(x) will have a pole at such a point, and using the formula for anti-derivative will be incorrect.

7) $h(x)= \sqrt x\ln(\sin x)$

Using the calculator of integrals, we obtain:

$\int\limits_{0.5}^3h(x)dx=-1.300$

$\int\limits_{3}^2h(x)dx=1.079$

$\int\limits_{7}^9 h(x)dx=-1.177$