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

Answer 1

1) $∫-4csc(x)cot(x)dx$

$-4(∫\frac{1}{sin\left(x\right)}\frac{\cos{\left(x\right)}}{\sin{\left(x\right)}}\ )\ dx$

$-4∫\frac{\cos{\left(x\right)}}{\sin^2{(x)}}\ dx$

let u = sin(x)

du = cos(x) dx

$-4∫\frac{du}{u^2}$

$\frac{4}{u}+c$

\frac{4}{\sin{\left(x\right)}}+c\

$4\csc{\left(x\right)}+c$

2) $∫-1\cot^2{\ \left(x\right)\ dx\ }$

$-1∫\frac{\cos^2{\left(x\right)}}{\sin^2{\left(x\right)}}\ dx$

$-1∫\ \frac{{1-\sin}^2{\left(x\right)}}{\sin^2{\left(x\right)}}dx$

$-1∫\ \frac{1}{\sin^2{\left(x\right)}}dx\ +\ 1\int dx$

let t = tan(x)

$dt = \ sec^2\left(x\right)\ dx$

$sin(x) = \frac{t}{\sqrt{1+t^2}}$

$\frac{1}{\sin^2{\left(x\right)}}=\frac{1+t^2}{t^2}$

$\cos{\left(x\right)}=\frac{1}{\sqrt{1+t^2}}$

$\sec^2{\left(x\right)}=1+t^2$

$dx=\frac{dt}{1+t^2}$

$-1∫\ \frac{1}{\sin^2{\left(x\right)}}dx+1\int dx=\ -1\int\frac{1+t^2}{t^2}\frac{dt}{1+t^2}+\int\frac{dt}{1+t^2}$

$-1∫\frac{dt}{t^2}\ +\tan^{-1}{\ }(t) + c2$

$\frac{1}{t}+ \tan^{-1}{\ }\left(t\right) + c$

$\frac{1}{\tan{\left(x\right)}}+\tan^{-1}{\ }\left(\tan{\left(x\right)}\right)+c$

$\cot{\left(x\right)}+x+c$

3) ∫(6x^{-3}+x^\frac{2}{3})\ dx\

$-3x^{-2}+\frac{3x^\frac{5}{3}}{5} + c$

4) $\int_{-1}^{3}{(10}e^x-4x\ )\ dx$

$(10e^x-2x^2)$

substituting the limits

$\left(10e^3-2\left(3^2\right)\right)-(10e^{-1}-2\left(-1\right)^2)$

$10\left(e^3-e^{-1}\right)-16$

5) $\int_{\frac{\pi}{2}}^{\frac{5\pi}{6}}\left(-3\cos{\left(x\right)}-4\sin{\left(x\right)}\ \right)dx$

$(-3\sin{\left(x\right)}+4\cos(x))$

substituting the limits,

$(-3\sin{\left(\frac{5\pi}{6}\right)}+4\cos{\left(\frac{5\pi}{6}\right)}\ )-(-3\sin{\left(\frac{\pi}{2}\right)}+4\cos(\frac{\pi}{2}))$

$-3\left(\frac{1}{2}\right)+4\left(-\frac{\sqrt3}{2}\right)+3(1)-4(0)$

$\frac{3-4\sqrt3}{2}$