MHB Quick Integral (U-substitution and partial fractions) Questions

[ardentmed]

Hey guys,

I'd really appreciate it if I could get some quick help for this problem set I'm working on.

For question one, I just did a quick u substitution for x^4 and managed to get x^4 * sin(x^4)+cos(x^4) + C.
For part b, I used integration by parts and took ln(4t) as u and the rest as the dv value. I got (2/3)t^(3/2)*ln(4t) - (4/9)t^(3/2) +C

For question two, I'm really doubting my response, especially for part b. I ended up proving the solution by taking u=(cos(x))^(n-1), expanding the n-1 when doing ibp, carrying over the n*(integral) and then dividing the other side by n.

Also, for question two part b, I'm almost clueless as to what I (thought) I did. I broke down cosx^4 to (cosx)^3 * sinx and (cosx)^2 sinx ^2 respectively. Then I changed sinx^2 to (1-cos2x)/2 via the double angle identity. Then I repeatedly performed ibp to get (-1/2)((cosx)^3)(sinx) + (3/4)cosxsinx - (3/4)x + c. Is that even remotely close to the answer?

 

[Prove It]

Hey guys,

I'd really appreciate it if I could get some quick help for this problem set I'm working on.

For question one, I just did a quick u substitution for x^4 and managed to get x^4 * sin(x^4)+cos(x^4) + C.
For part b, I used integration by parts and took ln(4t) as u and the rest as the dv value. I got (2/3)t^(3/2)*ln(4t) - (4/9)t^(3/2) +C

For question two, I'm really doubting my response, especially for part b. I ended up proving the solution by taking u=(cos(x))^(n-1), expanding the n-1 when doing ibp, carrying over the n*(integral) and then dividing the other side by n.

Also, for question two part b, I'm almost clueless as to what I (thought) I did. I broke down cosx^4 to (cosx)^3 * sinx and (cosx)^2 sinx ^2 respectively. Then I changed sinx^2 to (1-cos2x)/2 via the double angle identity. Then I repeatedly performed ibp to get (-1/2)((cosx)^3)(sinx) + (3/4)cosxsinx - (3/4)x + c. Is that even remotely close to the answer?

Finally, for question 7a, I used
1/2 * 4/8 [1f(x) + 2 + 2 + 2 + 1.. etc]
and computed 1.732865 as the answer.

For part b, I used 1/4, 3/4.. (etc.).. up to 15/4 for the midpoints, added the f(x) values of those points together, and divided by four. Oddly enough, I only got 0.893713 this time.

Finally, for question 7c, I used
1/3 * 4/8 [1f(x) + 4 + 2+ 4+ 2 + 4.. etc]
and computed 1.719693 as the answer, which is close to the trapezoidal approximation, but the midpoint approximation is what's throwing me off.

Thanks a ton, I really appreciate the help guys.

Well I can tell straight away your first question has at least one mistake in it, since $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( x^4 \right) = 4x^3 \end{align*}$, you are definitely out by a factor of 4.

When you substitute $\displaystyle \begin{align*} t = x^4 \implies \mathrm{d}t = 4x^3\,\mathrm{d}x \end{align*}$ you should have

$\displaystyle \begin{align*} \int{ x^7 \cos{ \left( x^4 \right) } \,\mathrm{d}x} &= \frac{1}{4} \int{ x^4 \cos{ \left( x^4 \right) } \, 4x^3 \, \mathrm{d}x } \\ &= \frac{1}{4} \int{ t\cos{ \left( t \right) } \, \mathrm{d}t } \end{align*}$

and now try integration by parts with $\displaystyle \begin{align*} u = t \implies \mathrm{d}u = \mathrm{d}t \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = \cos{ (t) } \,\mathrm{d}t \implies v = \sin{(t)} \end{align*}$...Your answer to (b) is correct.

 

[MarkFL]

Hey guys,

I'd really appreciate it if I could get some quick help for this problem set I'm working on...

Again, please post no more than two questions per thread. If a question contains multiple parts, then only post that one question in a thread.

When you post a large number of question in one thread, then the thread can become convoluted and hard to follow. Our rule regarding the limit of the number of questions per thread is designed to make MHB more efficient for everyone involved, for those posting questions, for those giving help, and for those doing searches.

 

[Prove It]

Hey guys,

I'd really appreciate it if I could get some quick help for this problem set I'm working on.

For question one, I just did a quick u substitution for x^4 and managed to get x^4 * sin(x^4)+cos(x^4) + C.
For part b, I used integration by parts and took ln(4t) as u and the rest as the dv value. I got (2/3)t^(3/2)*ln(4t) - (4/9)t^(3/2) +C

For question two, I'm really doubting my response, especially for part b. I ended up proving the solution by taking u=(cos(x))^(n-1), expanding the n-1 when doing ibp, carrying over the n*(integral) and then dividing the other side by n.

Also, for question two part b, I'm almost clueless as to what I (thought) I did. I broke down cosx^4 to (cosx)^3 * sinx and (cosx)^2 sinx ^2 respectively. Then I changed sinx^2 to (1-cos2x)/2 via the double angle identity. Then I repeatedly performed ibp to get (-1/2)((cosx)^3)(sinx) + (3/4)cosxsinx - (3/4)x + c. Is that even remotely close to the answer?

Finally, for question 7a, I used
1/2 * 4/8 [1f(x) + 2 + 2 + 2 + 1.. etc]
and computed 1.732865 as the answer.

For part b, I used 1/4, 3/4.. (etc.).. up to 15/4 for the midpoints, added the f(x) values of those points together, and divided by four. Oddly enough, I only got 0.893713 this time.

Finally, for question 7c, I used
1/3 * 4/8 [1f(x) + 4 + 2+ 4+ 2 + 4.. etc]
and computed 1.719693 as the answer, which is close to the trapezoidal approximation, but the midpoint approximation is what's throwing me off.

Thanks a ton, I really appreciate the help guys.

To do 2. a) which is to come up with the reduction formula $\displaystyle \begin{align*} \int{ \cos^n{(x)}\,\mathrm{d}x} = \frac{1}{n}\cos^{n-1}{(x)} \sin{(x)} + \frac{n - 1}{n} \int{ \cos^{n-2}{(x)}\,\mathrm{d}x} \end{align*}$, we use integration by parts with $\displaystyle \begin{align*} u = \cos^{n - 1}{(x)} \implies \mathrm{d}u = -\left( n - 1 \right) \sin{(x)} \cos^{n-2}{(x)} \,\mathrm{d}x \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = \cos{(x)} \,\mathrm{d}x \implies v = \sin{(x)} \end{align*}$, then

$\displaystyle \begin{align*} I &= \int{ \cos^n{(x)}\,\mathrm{d}x} \\ I &= \sin{(x)}\cos^{n-1}{(x)} - \int{ - \left( n - 1 \right) \sin^2{(x)} \cos^{n-2}{(x)}\,\mathrm{d}x } \\ I &= \sin{(x)}\cos^{n-1}{(x)} + \left( n - 1 \right) \int{ \sin^2{(x)}\cos^{n-2}{(x)} \,\mathrm{d}x} \\ I &= \sin{(x)}\cos^{n-1}{(x)} + \left( n - 1 \right) \int{ \left[ 1 - \cos^2{(x)} \right] \cos^{n - 2 }{(x)} \, \mathrm{d}x } \\ I &= \sin{(x)} \cos^{n - 1}{(x)} + \left( n - 1 \right) \int{ \cos^{n - 2}{(x)} - \cos^n{(x)}\,\mathrm{d}x} \\ I &= \sin{(x)}\cos^{n-1}{(x)} + \left( n - 1 \right) \int{ \cos^{n-2}{(x)}\,\mathrm{d}x} - \left( n - 1 \right) \int{ \cos^n{(x)}\,\mathrm{d}x} \\ I &= \sin{(x)} \cos^{n-1}{(x)} + \left( n - 1 \right) \int{ \cos^{n-2}{(x)}\,\mathrm{d}x} - \left( n - 1 \right) I \\ I + \left( n - 1 \right) I &= \sin{(x)}\cos^{n-1}{(x)} + \left( n - 1 \right) \int{ \cos^{n-2}{(x)}\,\mathrm{d}x} \\ n\,I &= \sin{(x)}\cos^{n-1}{(x)} + \left( n - 1 \right) \int{ \cos^{n-2}{(x)}\,\mathrm{d}x} \\ I &= \frac{1}{n}\sin{(x)}\cos^{n-1}{(x)} + \frac{n - 1}{n} \int{ \cos^{n-2}{(x)} \, \mathrm{d}x} \end{align*}$

For part (b) you're literally just replacing n with 4 here, then in the resulting integral using the same formula replacing n with 2...

$\displaystyle \begin{align*} \int{ \cos^4{(x)} \, \mathrm{d}x} &= \frac{1}{4} \sin{(x)} \cos^3{(x)} + \frac{3}{4} \int{ \cos^2{(x)} \, \mathrm{d}x} \\ &= \frac{1}{4}\sin{(x)}\cos^3{(x)} + \frac{3}{4} \left[ \frac{1}{2} \sin{(x)}\cos{(x)} + \frac{1}{2} \int{ \cos^0{(x)}\,\mathrm{d}x} \right] \\ &= \frac{1}{4}\sin{(x)}\cos^3{(x)} + \frac{3}{8} \sin{(x)}\cos{(x)} + \frac{3}{8}\int{ 1\,\mathrm{d}x} \\ &= \frac{1}{4}\sin{(x)}\cos^3{(x)} + \frac{3}{8}\sin{(x)}\cos{(x)} + \frac{3}{8}x + C\end{align*}$

 

[soroban]

Hello, ardentmed!

Here is some help . . .

(a)\;\;I \;=\;\int x^7\cos(x^4)\,dx

By parts: $\:\begin{Bmatrix}u &=& x^4 && dv &=& x^3\cos(x^4)\,dx \\ du &=& 4x^3\,dx && v &=& \frac{1}{4}\sin(x^4) \end{Bmatrix}$

$\displaystyle I \;=\;\tfrac{1}{4}x^4\sin(x^4) - \int x^3\sin(x^4)\,dx $

$ I \;=\;\frac{1}{4}x^4\sin(x^4) + \frac{1}{4}\cos(x^4) + C$

(b)\;\;I \;=\;\int \sqrt{t}\,\ln(4t)\,dt

By parts: $\:\begin{Bmatrix}u &=& \ln(4x) && dv &=& x^{\frac{1}{2}}dx \\ du &=& \frac{dx}{x} && v &=& \frac{2}{3}x^{\frac{3}{2}} \end{Bmatrix}$

$\displaystyle I \;=\;\tfrac{2}{3}x^{\frac{3}{2}}\ln(4x) - \tfrac{2}{3}\int x^{\frac{1}{2}}\,dx $

$I \;=\;\frac{2}{3}x^{\frac{3}{2}}\ln(4x) - \frac{4}{9}x^{\frac{3}{2}} + C$

$I \;=\;\frac{2}{9}x^{\frac{3}{2}}\big[3\ln(4x) - 2\big] + C$

 

[ardentmed]

Again, please post no more than two questions per thread. If a question contains multiple parts, then only post that one question in a thread.

When you post a large number of question in one thread, then the thread can become convoluted and hard to follow. Our rule regarding the limit of the number of questions per thread is designed to make MHB more efficient for everyone involved, for those posting questions, for those giving help, and for those doing searches.

Duly noted, I edited the post and posted question seven in a new thread. Sorry about that.