Integrals Involving Transcendental Functions

Evaluate the following integrals.

  1. \displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx
  2. \displaystyle \int\frac{2^{x}\: dx}{\left(2^{x}+1\right)\sqrt{4^{x}+2^{x+1}-8}}
  3. \displaystyle \int\frac{dx}{x+4\sqrt{x}+13}

Solution

We use the substitution rule in the following computations.

For the first integral, let u =\ln\left(\sin x\right). Then du=\dfrac{1}{\sin x}\cdot\cos x\: dx=\cot x\: dx. Moreover, if x=\dfrac{\pi}{6}, u=\ln\left(\dfrac{1}{2}\right)=-\ln2 and if x=\dfrac{\pi}{4}, u=\ln\left(\dfrac{\sqrt{2}}{2}\right)=-\dfrac{1}{2}\ln2. Thus,

\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\ln^{2}\left(\sin x\right)\cot x\: dx

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx } = \int_{-\ln2}^{-\frac{1}{2}\ln2}u^{2}\: du

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx } = \left.\frac{u^{3}}{3}\right|_{-\ln2}^{-\frac{1}{2}\ln2}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx } = \frac{1}{3}\left(-\frac{1}{8}\ln^{3}2+\ln^{3}2\right)

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx } = \frac{7}{24}\ln^{3}2

In the second integral, note that

\displaystyle 4^{x}+2^{x+1}-8=2^{2x}+2\cdot2^{x}+1-9=\left(2^{x}+1\right)^{2}-3^{2}.

So we suspect that the given integral yields the inverse secant function. Indeed, if we make the substitution u=2^{x}+1, then du=2^{x}\ln2dx and thus

\displaystyle \int\frac{2^{x}\: dx}{\left(2^{x}+1\right)\sqrt{4^{x}+2^{x+1}-8}} = \frac{1}{\ln2}\int\frac{du}{u\sqrt{u^{2}-3^{2}}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{2^{x}\:dx}{\left( 2^{x}+1\right) \sqrt{4^{x}+2^{x+1}-8}}} = \frac{1}{\ln2}\cdot\frac{1}{3} \sec^{-1} \left( \frac{u}{3}\right) +C

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{2^{x}\:dx}{\left( 2^{x}+1\right) \sqrt{4^{x}+2^{x+1}-8}}} = \frac{1}{3\ln2}\sec^{-1} \left( \frac{2^{x}+1}{3}\right) +C.

For the last integral, let’s make the substitution u=\sqrt{x}. Then du=\dfrac{1}{2\sqrt{x}}\, dx or dx=2udu. So we have

\displaystyle \int\frac{dx}{x+4\sqrt{x}+13} = \int\frac{2u\, du}{u^{2}+4u+13}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{dx}{x+4\sqrt{x}+13}} = \int\frac{2u+4}{u^{2}+4u+13}\,du-\int\frac{4\,du}{u^{2}+4u+13}.

Observe that the first part above yields the natural logarithmic function while the second part involves the inverse tangent function. In fact, if we let v=u^{2}+4u+13, then dv=\left(2u+4\right)du. Thus,

\displaystyle \int\frac{2u+4}{u^{2}+4u+13}\, du = \int\frac{dv}{v}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{2u+4}{u^{2}+4u+13}\, du} = \ln\left|v\right|+C_{1}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{2u+4}{u^{2}+4u+13}\, du} = \ln\left(u^{2}+4u+13\right)+C_{1}.

Meanwhile,

\displaystyle \int\frac{4\, du}{u^{2}+4u+13} = 4\int\frac{du}{\left(u+2\right)^{2}+3^{2}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{4\, du}{u^{2}+4u+13}} = \frac{4}{3}\tan^{-1}\left(\frac{u+2}{3}\right)+C_{2}.

Hence, the third integral yields

\displaystyle \ln\left(x+4\sqrt{x}+13\right)-\frac{4}{3}\tan^{-1}\left(\frac{\sqrt{x}+2}{3}\right)+C.

This entry was posted on Sunday, March 23rd, 2008 at 2:37 am and is filed under Calculus. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

4 Responses to “Integrals Involving Transcendental Functions”

  1. rohedi Says:
    December 10, 2008 at 4:51 am | Reply

    If the integral in the last problem is converted into ordinary differential equation (ODE) , then we will have

    dx/dt = x + 4sqrt(x) + 13

    My question, how to solve the above ODE without of both changing variable and separating variable

  2. Nadya Fermega Says:
    December 25, 2008 at 2:12 am | Reply

    Rohedi, to solve your problem please visit to this address http://eqworld.ipmnet.ru/forum/viewtopic.php?f=4&t=41.

  3. Rohedi Says:
    April 10, 2009 at 1:57 pm | Reply

    Hi All,

    In the following link:

    http://eqworld.ipmnet.ru/forum/viewtopic.php?f=3&t=148,

    there are a post that discusses a topic related to “transcendental word” that is the general form of Pi exact formula. Maybe useful for you, please visit to the link.

  4. Nadya Fermega Says:
    May 4, 2009 at 6:11 am | Reply

    Pi(Phi) formula? Ohh Nadya must help daddy Rohedi about the nice number for presenting the pi exact formula in form of Phi golden ratio that posted at http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=157.

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