integral calculus problem lists

nephi39 · January 11th, 2012, 08:00 AM

Guys our examination scheduled last week of january..

and our teacher gave us 30 home work problems last week to help us calculus for our upcoming examination...

and tomorrow is the deadline of submission for this 30 problems...

can u help me solve the remaining 5 problems.. i just rely on examples to solve problems,
i really dont know how to answer those 5, dont have any idea, please help me...

1. integral of 4e raise to 3lnx

2. integral 1-cos 2r divided by 1+cos 2r

3. integral tan x divided by 1-tan^2 x

4. integral sinh 4x divided by cosh 2x

5. integral x sech^2 xdx

i try to answer number 5 and this is how i got my answer
I let u = x then du = dx

integral of x sech^2 u dx i cancel out x and left with integral sech^2 udx and integrate it and got this answer

tanh x + c << this is my final answer, Guys Am i Correct with this or wrong

can u help me please.. its time to sleep here but tomorrow is the deadline for all of this

Guys please help me here....

can u answer and put some explanation so i will know how to solve problems like this in our examination

thank you very much...

The Chaz · January 11th, 2012, 11:04 AM

1. Use these:
e^(ln a) = a
b ln a = ln a^b

Then you have (just) the integral of 4x^3. It's a trick question, in a sense.

2. Appropriate algebraic/trig manipulations/identities should give you a simpler integrand...

3. Same. The denominator is -1*sec^2(x), if memory serves.

MarkFL · January 11th, 2012, 11:47 AM

1.) $\int 4e^{3\ln x}\,dx=4\int e^{\ln x^3}\,dx=4\int x^3\,dx=x^4+C$

The trick here is the use the identity $a^{\log_a b}=b$ .

2.) $\int \frac{1-\cos(2r)}{1+\cos(2r)}\,dr$

Using double-angle identities for cosine: $\cos$2\theta$\,=\,2\cos^2$\theta$-1\,=\,1-2\sin^2$\theta$$

$\int \frac{2\sin^2r}{2\cos^2r}\,dr=\int \tan^2r\,dr$

Using a Pythagorean identity $\tan^2\theta=\sec^2\theta-1$ we have:

$\int \sec^2r-1\,dr=\tan r-r+C$

3.) $\int \frac{\tan x}{1-\tan^2x}\,dx$

Using a double-angle identity for tangent: $\tan$2\theta$=\frac{2\tan\theta}{1-\tan^2\theta}$ we have:

$\frac{1}{2}\int \tan(2x)\,dx$

Let $u=\cos(2x)\:\therefore\:du=-2\sin(2x)$ and we have:

$-\frac{1}{4}\int \frac{1}{u}\,du=-\frac{1}{4}\ln|u|+C=-\frac{1}{4}\ln|\cos(2x)|+C$

4.) $\int \frac{\sinh(4x)}{\cosh(2x)}\,dx$

Using a double-argument identity for the hyperbolic sine function: $\sinh(2x)=2\sinh(x)\cosh(x)$ we have:

$\int \frac{2\sinh(2x)\cosh(2x)}{\cosh(2x)}\,dx=2\int \sinh(2x)\,dx=\cosh(2x)+C$

5.) $\int x\cdot\text{sech}^2(x)\,dx$

Using integration by parts:

Let $u=x\:\therefore\:du=dx$ and $dv=\text{sech}^2(x)\,dx\:\therefore\:v=\tanh(x)$ and we have:

$x\tanh(x)-\int \tanh(x)\,dx=x\tanh(x)-\ln$\cosh(x)$+C$

nephi39 · January 11th, 2012, 02:25 PM

sir Mark

regarding number 1 and 3 can u add some explanation sir, im not good in calculus but im trying my best

regarding number 1, is it only x^4+c or i can answer e x^4 +c

because in our exponential function lectures all examples given by my instructor always come up with an answer with "e"

and in number 3 problem

1/2 integral of tan(2x)dx why u=cos2x

thank you very much.. its help me a lot... u too are so great

tnx for ur patience on slow mind students like me...

nephi39 · January 11th, 2012, 02:33 PM

sir mark.. i already understand number 3

after 1/2 integral tan(2x)dx

u change it to (1/2) sin(2x) divided by cos(2x) so u= cos(2x)

tnx again sir mark,

MarkFL · January 11th, 2012, 04:30 PM

Glad you figured out number 3, I did indeed use tan(x)=sin(x)/cos(x).

For number 1, we have $4\int e^{3\ln x}\,dx$

Now, if we choose to solve this without simplifying first using the above mentioned identity, we could choose to let:

$u=3\ln x\:\therefore\:du=\frac{3}{x}\,dx$ and so we could write:

$\frac{4}{3}\int x\cdot e^{3\ln x}\frac{3}{x}\,dx$ so now we have:

$\frac{4}{3}\int e^{\frac{u}{3}}\cdot e^u\,du=\int e^{\frac{4}{3}u}$\frac{4}{3}\,du$$

Let $v=\frac{4}{3}u\:\therefore\:dv=\frac{4}{3}\,du$ giving us:

$\int e^v\,dv=e^v+C=e^{\frac{4}{3}u}+C=e^{\frac{4}{3}\cd ot3\ln(x)}+C=e^{4\ln(x)}+C=e^{\ln$x^4$}+C$

Now we are back to using the above identity again to simplify:

$4\int e^{3\ln x}\,dx=x^4+C$

As you see the answer can be expressed as $e^{f(x)}+C$ but we really should simplify it.

We cannot use $ex^4+C$ since $\frac{d}{dx}$ex^4+C$=4ex^3\ne4e^{3\ln(x)}$ .

A word about the identity $a^{\log_a(b)}=b$

Let:

$y=a^{\log_a(b)}$

Convert from exponential to logarithmic form:

$\log_a(y)=\log_a(b)$

Taking the anti-log of both sides gives:

$y=b$ and thus:

$b=a^{\log_a(b)}$

nephi39 · January 11th, 2012, 04:45 PM

thank you thank you very much sir..
surely it will helps me for my examination next next week..

thank you very much for your time,

January 11th, 2012, 08:00 AM	#1
nephi39 Newbie Joined: Dec 2011 Posts: 25 Thanks: 0	integral calculus list of problems Guys our examination scheduled last week of january.. and our teacher gave us 30 home work problems last week to help us calculus for our upcoming examination... and tomorrow is the deadline of submission for this 30 problems... can u help me solve the remaining 5 problems.. i just rely on examples to solve problems, i really dont know how to answer those 5, dont have any idea, please help me... 1. integral of 4e raise to 3lnx 2. integral 1-cos 2r divided by 1+cos 2r 3. integral tan x divided by 1-tan^2 x 4. integral sinh 4x divided by cosh 2x 5. integral x sech^2 xdx i try to answer number 5 and this is how i got my answer I let u = x then du = dx integral of x sech^2 u dx i cancel out x and left with integral sech^2 udx and integrate it and got this answer tanh x + c << this is my final answer, Guys Am i Correct with this or wrong can u help me please.. its time to sleep here but tomorrow is the deadline for all of this Guys please help me here.... can u answer and put some explanation so i will know how to solve problems like this in our examination thank you very much...
$nephi39 is offline$

January 11th, 2012, 11:04 AM	#2
The Chaz Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4	Re: integral calculus list of problems 1. Use these: e^(ln a) = a b ln a = ln a^b Then you have (just) the integral of 4x^3. It's a trick question, in a sense. 2. Appropriate algebraic/trig manipulations/identities should give you a simpler integrand... 3. Same. The denominator is -1*sec^2(x), if memory serves.
$The Chaz is offline$

January 11th, 2012, 11:47 AM	#3
MarkFL Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs	Re: integral calculus list of problems 1.) $\int 4e^{3\ln x}\,dx=4\int e^{\ln x^3}\,dx=4\int x^3\,dx=x^4+C$ The trick here is the use the identity $a^{\log_a b}=b$ . 2.) $\int \frac{1-\cos(2r)}{1+\cos(2r)}\,dr$ Using double-angle identities for cosine: $\cos\(2\theta\)\,=\,2\cos^2\(\theta\)-1\,=\,1-2\sin^2\(\theta\)$ $\int \frac{2\sin^2r}{2\cos^2r}\,dr=\int \tan^2r\,dr$ Using a Pythagorean identity $\tan^2\theta=\sec^2\theta-1$ we have: $\int \sec^2r-1\,dr=\tan r-r+C$ 3.) $\int \frac{\tan x}{1-\tan^2x}\,dx$ Using a double-angle identity for tangent: $\tan\(2\theta\)=\frac{2\tan\theta}{1-\tan^2\theta}$ we have: $\frac{1}{2}\int \tan(2x)\,dx$ Let $u=\cos(2x)\:\therefore\:du=-2\sin(2x)$ and we have: $-\frac{1}{4}\int \frac{1}{u}\,du=-\frac{1}{4}\ln\|u\|+C=-\frac{1}{4}\ln\|\cos(2x)\|+C$ 4.) $\int \frac{\sinh(4x)}{\cosh(2x)}\,dx$ Using a double-argument identity for the hyperbolic sine function: $\sinh(2x)=2\sinh(x)\cosh(x)$ we have: $\int \frac{2\sinh(2x)\cosh(2x)}{\cosh(2x)}\,dx=2\int \sinh(2x)\,dx=\cosh(2x)+C$ 5.) $\int x\cdot\text{sech}^2(x)\,dx$ Using integration by parts: Let $u=x\:\therefore\:du=dx$ and $dv=\text{sech}^2(x)\,dx\:\therefore\:v=\tanh(x)$ and we have: $x\tanh(x)-\int \tanh(x)\,dx=x\tanh(x)-\ln\(\cosh(x)\)+C$
$MarkFL is offline$

January 11th, 2012, 02:25 PM	#4
nephi39 Newbie Joined: Dec 2011 Posts: 25 Thanks: 0	Re: integral calculus list of problems sir Mark regarding number 1 and 3 can u add some explanation sir, im not good in calculus but im trying my best regarding number 1, is it only x^4+c or i can answer e x^4 +c because in our exponential function lectures all examples given by my instructor always come up with an answer with "e" and in number 3 problem 1/2 integral of tan(2x)dx why u=cos2x thank you very much.. its help me a lot... u too are so great tnx for ur patience on slow mind students like me...
$nephi39 is offline$

January 11th, 2012, 02:33 PM	#5
nephi39 Newbie Joined: Dec 2011 Posts: 25 Thanks: 0	Re: integral calculus list of problems sir mark.. i already understand number 3 after 1/2 integral tan(2x)dx u change it to (1/2) sin(2x) divided by cos(2x) so u= cos(2x) tnx again sir mark,
$nephi39 is offline$

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January 11th, 2012, 04:30 PM	#6
MarkFL Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs	Re: integral calculus list of problems Glad you figured out number 3, I did indeed use tan(x)=sin(x)/cos(x). For number 1, we have $4\int e^{3\ln x}\,dx$ Now, if we choose to solve this without simplifying first using the above mentioned identity, we could choose to let: $u=3\ln x\:\therefore\:du=\frac{3}{x}\,dx$ and so we could write: $\frac{4}{3}\int x\cdot e^{3\ln x}\frac{3}{x}\,dx$ so now we have: $\frac{4}{3}\int e^{\frac{u}{3}}\cdot e^u\,du=\int e^{\frac{4}{3}u}\(\frac{4}{3}\,du\)$ Let $v=\frac{4}{3}u\:\therefore\:dv=\frac{4}{3}\,du$ giving us: $\int e^v\,dv=e^v+C=e^{\frac{4}{3}u}+C=e^{\frac{4}{3}\cd ot3\ln(x)}+C=e^{4\ln(x)}+C=e^{\ln\(x^4\)}+C$ Now we are back to using the above identity again to simplify: $4\int e^{3\ln x}\,dx=x^4+C$ As you see the answer can be expressed as $e^{f(x)}+C$ but we really should simplify it. We cannot use $ex^4+C$ since $\frac{d}{dx}\(ex^4+C\)=4ex^3\ne4e^{3\ln(x)}$ . A word about the identity $a^{\log_a(b)}=b$ Let: $y=a^{\log_a(b)}$ Convert from exponential to logarithmic form: $\log_a(y)=\log_a(b)$ Taking the anti-log of both sides gives: $y=b$ and thus: $b=a^{\log_a(b)}$
$MarkFL is offline$

January 11th, 2012, 04:45 PM	#7
nephi39 Newbie Joined: Dec 2011 Posts: 25 Thanks: 0	Re: integral calculus list of problems thank you thank you very much sir.. surely it will helps me for my examination next next week.. thank you very much for your time,
$nephi39 is offline$

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