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more definite integrals

sbhatnagar

Active member
Jan 27, 2012
95
Fun! Fun! Fun! Here are more entertaining problems:

1.\( \displaystyle \int_{2}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}dx\)

2.\( \displaystyle \int_{\sqrt{\ln(2)}}^{\sqrt{\ln(3)}}\frac{x \sin^2(x)}{\sin(x^2)+\sin(\ln(6)-x^2)}dx\)

3.\( \displaystyle \int_{-\pi/2}^{\pi/2}\frac{1}{1+4563^x}\frac{\sin^{6792}(x)}{\sin^{6792}(x)+\cos^{6792}(x)}dx\)

4.\( \displaystyle \int_{0}^{2} \frac{dx}{(17+8x-4x^2)(e^{6(1-x)}+1)} \)


These problems are very simple only if you know the right trick.
 
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Sherlock

Member
Jan 28, 2012
59
Fun! Fun! Fun! Here are more entertaining problems:

1.\( \displaystyle \int_{2}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}dx\)
Put $x \mapsto 6-x$ then we've $ \begin{aligned} I & = \int_{2}^{4}\frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}\;{dx} = \int_{2}^{4}\frac{\sqrt{\ln(3+x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}}\;{dx}\end{aligned}$.
Add these together and we've $\begin{aligned}2I = \int_{2}^{4}\frac{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}\;{dx} = \int_{2}^{4}\;{dx} = 2.\end{aligned}$ Therefore $I = 1$.
 
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ThePerfectHacker

Well-known member
Jan 26, 2012
236
Whenever I teach Calculus I assign the following homework problem (same idea):

$$ \int_0^2 \frac{ \sin^{2012} \left( \log (1+x) \right) }{ \sin^{2012} \left( \log(1+x) \right) + \sin^{2012} \left( \log(3-x)\right) } ~ dx $$
 

Sherlock

Member
Jan 28, 2012
59

3.\( \displaystyle \int_{-\pi/2}^{\pi/2}\frac{1}{1+4563^x}\frac{\sin^{6792}(x)}{\sin^{6792}(x)+\cos^{6792}(x)}dx\)
Put $x \mapsto -x$ then we've $\begin{aligned} I = \int_{-\pi/2}^{\pi/2}\frac{1}{1+4563^{x}}\frac{\sin^{6792}{x}}{\sin^{6792}{x}+\cos^{6792}{x}}\;{dx} = \int_{-\pi/2}^{\pi/2}\frac{1}{1+4563^{-x}}\frac{\sin^{6792}{x}}{\sin^{6792}{x}+\cos^{6792}{x}}\;{dx}.\end{aligned}$
Add them to get $\begin{aligned}2I = \int_{-\pi/2}^{\pi/2}\bigg(\frac{1}{1+4563^x}+\frac{1}{1+4563^{-x}}\bigg)\frac{\sin^{6792}{x}}{\sin^{6792}{x}+\cos^{6792}{x}}\;{dx} = 2\int_{0}^{\pi/2}\frac{\sin^{6792}{x}}{\sin^{6792}{x}+\cos^{6792}{x}}\;{dx}.\end{aligned}$
Let $x \mapsto \frac{\pi}{2}-x$ then $\begin{aligned}I = \int_{0}^{\pi/2}\frac{\sin^{6792}(\frac{\pi}{2}-x)}{\sin^{6792}(\frac{\pi}{2}-x)+\cos^{6792}(\frac{\pi}{2}-x)}\;{dx} = \int_{0}^{\pi/2}\frac{\cos^{6792}{x}}{\sin^{6792}{x}+\cos^{6792}{x}}\;{dx} \end{aligned}$, add them to get:
$\begin{aligned}2I = \int_{0}^{\pi/2}\frac{\sin^{6792}{x}}{\sin^{6792}{x}+\cos^{6792}{x}}\;{dx}+\int_{0}^{\pi/2}\frac{\cos^{6792}{x}}{\sin^{6792}{x}+\cos^{6792}{x}}\;{dx} = \int_{0}^{\pi/2}\frac{\sin^{6792}{x}+\cos^{6792}{x}}{\sin^{6792}{x}+\cos^{6792}{x}}\;{dx} = \frac{\pi}{2}. \end{aligned}$ Thus $\begin{aligned}I = \frac{\pi}{4}.\end{aligned}$

---------- Post added at 05:07 PM ---------- Previous post was at 04:54 PM ----------

Whenever I teach Calculus I assign the following homework problem (same idea):

$$ \int_0^2 \frac{ \sin^{2012} \left( \log (1+x) \right) }{ \sin^{2012} \left( \log(1+x) \right) + \sin^{2012} \left( \log(3-x)\right) } ~ dx $$

Let $x \mapsto 2-x$ then
$ \begin{aligned} I= \int_0^2 \frac{ \sin^{2012} \left( \log (1+x) \right) }{ \sin^{2012} \left( \log(1+x) \right) + \sin^{2012} \left( \log(3-x)\right) } \;{dx} = \int_0^2 \frac{ \sin^{2012} \left( \log (3-x) \right) }{ \sin^{2012} \left( \log(1+x) \right) + \sin^{2012} \left( \log(3-x)\right) } \;{dx} \end{aligned}$

Therefore
$ \begin{aligned} 2I = \int_0^2 \frac{ \sin^{2012} \left( \log (1+x) \right) }{ \sin^{2012} \left( \log(1+x) \right) + \sin^{2012} \left( \log(3-x)\right) } \;{dx} + \int_0^2 \frac{ \sin^{2012} \left( \log (3-x) \right) }{ \sin^{2012} \left( \log(1+x) \right) + \sin^{2012} \left( \log(3-x)\right) } \;{dx} \end{aligned}$, thus:
$ \begin{aligned}2I = \int_{0}^{2}\frac{\sin^{2012} \left( \log (1+x) \right) +\sin^{2012} \left( \log (3-x) \right)}{​\sin^{2012} \left( \log(1+x) \right) + \sin^{2012} \left( \log(3-x)\right) }\;{dx} = \int_{0}^{2}\;{dx} = 2. \end{aligned}$ Hence $2I = 2$ and so $I = 1$.

What I find interesting is that the trick also applies to products and sums because the index can be shifted in the same way:


$\displaystyle \int_{a}^{b}f(x)\;{dx} = \int_{a}^{b}f(a+b-x)\;{dx}$,
$\displaystyle \sum_{a \le k \le b}f(k) = \sum_{a \le k \le b}f(a+b-k) $ and $\displaystyle ~ \prod_{a \le k \le b}f(k) = \prod_{a \le k \le b}f(a+b-k). $

It's amazing! So you can also create a simple but monstrous looking sum or product when you are teaching these topics too.
 
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Sherlock

Member
Jan 28, 2012
59
I think you had a typo on this one; I fixed it so that it's similar to other integrals.


2.\( \displaystyle \int_{\sqrt{\ln(2)}}^{\sqrt{\ln(3)}}\frac{x \sin(x^2)}{\sin(x^2)+\sin(\ln(6)-x^2)}dx\)
Let $\displaystyle t = x^2$ then $\displaystyle I = \frac{1}{2}\int_{\ln{2}}^{\ln{3}}\frac{ \sin{t}}{\sin{t}+\sin(\ln{6}-t)}\;{dt}$. Now put $t\mapsto \ln{6}-t$, and we've $\displaystyle I = \frac{1}{2}
\int_{\ln{2}}^{\ln{3}}\frac{\sin(\ln{6}-t)}{\sin{t}+\sin(\ln{6}-t)}\;{dt} $, therefore:
$\displaystyle 2I = \frac{1}{2}\int_{\ln{2}}^{\ln{3}}\frac{ \sin{t}}{\sin{t}+\sin(\ln{6}-t)}\;{dt}+\frac{1}{2}\int_{\ln{2}}^{\ln{3}}\frac{ \sin( \ln{6}-t)}{\sin{t}+\sin(\ln{6}-t)}\;{dt} = \frac{1}{2}\int_{\ln{2}}^{\ln{3}}\frac{\sin{t}+ \sin(\ln{6}-t)}{\sin{t}+\sin(\ln{6}-t)}\;{dt} = \frac{1}{2}\ln\left(\frac{3}{2}\right).$ So $\displaystyle I = \frac{1}{4}\ln\left(\frac{3}{2}\right).$

---------- Post added at 11:25 PM ---------- Previous post was at 10:40 PM ----------

4.\( \displaystyle \int_{0}^{2} \frac{dx}{(17+8x-4x^2)(e^{6(1-x)}+1)} \)
Let $x\mapsto 2-x$ then $\displaystyle I = \int_{0}^{2}\frac{1}{(17+8x-4x^2)(e^{6(1-x)}+1)}\;{dx} = \int_{0}^{2}\frac{1}{(17+8x-4x^2)(e^{-6(1-x)}+1)}\;{dx}$.
Thus $\displaystyle 2I = \int_{0}^{2}\bigg(\frac{1}{e^{6(1-x)}+1}+\frac{1}{e^{-6(1-x)}+1}\bigg)\frac{1}{(17+8x-4x^2)}\;{dx} = \int_{0}^{2}\frac{1}{17+8x-4x^2}\;{dx},$ therefore:
$\displaystyle 2I = \int_{0}^{2}\frac{1}{2\sqrt{21}(2x+\sqrt{21}-2)}+\frac{1}{2\sqrt{21}(-2x+\sqrt{21}+2)}\;{dx} = \frac{1}{4\sqrt{21}}\ln\bigg|\frac{\sqrt{21}-2+2x}{\sqrt{21}+2-2x}\bigg|_{0}^{2} = \frac{1}{2\sqrt{21}}\ln\bigg|\frac{\sqrt{21}+2}{ \sqrt{21}-2}\bigg|$

Therefore $\displaystyle I = \frac{1}{4\sqrt{21}}\ln\bigg|\frac{\sqrt{21}+2}{ \sqrt{21}-2}\bigg|. $ There goes the last one. I had a field day with these integrals today. Thanks! :]


 
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