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 January 11th, 2012, 11:04 AM #2 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.
 January 11th, 2012, 11:47 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 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$
 January 11th, 2012, 02:25 PM #4 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...
 January 11th, 2012, 02:33 PM #5 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,
 January 11th, 2012, 04:30 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 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: $b=a^{\log_a(b)}$
 January 11th, 2012, 04:45 PM #7 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,

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