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I had to evaluate this integral .

$$ \int_0^{\pi/2} \frac{\cos x}{3 \cos x + \sin x} \, dx $$

Here is how I proceeded

Dividing $N^r$ And $D^r$ by $\cos^3 x$

$$ \int_0^{\pi/2} \frac{ \sec^2 x}{3 \sec^2 x + \tan x \sec^2 x}\, dx \\ $$

Substituting $\tan x = t$

$$ \int_0^\infty \frac{ 1 }{(1+t^2)(t+3)} \, dt \\ $$

Then by using Partial Fractions , I got the answer as

$$ \frac{1}{10} \log (t+3) - \frac{1}{20} \log (t^2 + 1) + \frac{3}{10} \arctan (t) \biggr|_{0}^{\infty} $$

But while substituting the limits , the answer comes out be be infinity which is wrong .

Is there any mistake in my approach ??

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Actually, you should have obtained$$\int\frac1{(1+t^2)(t+3)}\,\mathrm dt=\frac3{10}\arctan(t)+\frac1{10}\log(3+t)-\frac1{20}\log(1+t^2).$$Now, note that\begin{align}\frac1{10}\log(3+t)-\frac1{20}\log(1+t^2)&=\frac1{20}\left(\log\bigl((3+t)^2\bigr)-\log(1+t^2)\right)\\&=\frac1{20}\log\left(\frac{(3+t)^2}{1+t^2}\right).\end{align}

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    $\begingroup$ But still while substituting $\infty$ , Doesn't the answer become $\log \infty$ which is $\infty$ , so how do we get finite term ? $\endgroup$ – user589548 Oct 4 '18 at 1:56
  • $\begingroup$ No. Since $\lim_{t\to\infty}\frac{(3+t)^2}{1+t^2}=1$, if you replace $t$ by $\infty$ in $\log\left(\frac{(3+t)^2}{1+t^2}\right)$, you get $\log(1)$, which is $0$. $\endgroup$ – José Carlos Santos Oct 4 '18 at 5:37
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    $\begingroup$ And suppose we don't combine $\frac{1}{10} \log {3+t}$ and $\frac{1}{20} \log {1+t^2}$ , Then won't we be able to substitute $\infty$ ??? $\endgroup$ – user589548 Oct 4 '18 at 9:44
  • $\begingroup$ Then you get an indeterminate, which is $\infty-\infty$. $\endgroup$ – José Carlos Santos Oct 4 '18 at 9:46
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    $\begingroup$ BTW is there any other way of solving this question from start ? $\endgroup$ – user589548 Oct 4 '18 at 9:50
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A simple method to avoid using partial fractions and improper integrals. Let $$ A=\int_0^{\pi/2} \frac{\cos x}{3 \cos x + \sin x} \, dx, B=\int_0^{\pi/2} \frac{\sin x}{3 \cos x + \sin x} \, dx. $$ Clearly $$ 3A+B=\frac{\pi}{2}. \tag{1}$$ Noting that $$ A=\int_0^{\pi/2} \frac{1}{3 \cos x + \sin x} \, d\sin x, B=-\int_0^{\pi/2} \frac{1}{3 \cos x + \sin x} \, d\cos x $$ it is easy to see $$ -3B+A=\int_0^{\pi/2} \frac{1}{3 \cos x + \sin x} \, d(3\cos x+\sin x)=\ln(3\cos x+\sin x)\bigg|_0^{\frac\pi2}=-\ln 3.\tag{2} $$ From (1) and (2), one has $$A=\frac{3\pi}{20}-\frac{1}{10}\ln3.$$

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$$\int_0^\infty \frac{ 1 }{(1+t^2)(t+3)} \, dt =\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{10}}}{\displaystyle\int_0^\infty}\dfrac{1}{t+3}\,\mathrm{d}t-\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{10}}}{\displaystyle\int_0^\infty}\dfrac{t-3}{t^2+1}\,\mathrm{d}t\\=\dfrac{\ln\left(\left|t+3\right|\right)}{10}-\dfrac{\ln\left(t^2+1\right)}{20}+\dfrac{3\arctan\left(t\right)}{10}\biggr|_{0}^{\infty}\\$$

$$=\frac1{20}\ln\frac{(t+3)^2}{1+t^2}+\dfrac{3\arctan\left(t\right)} {10}\biggr|_{0}^{\infty}$$

$$\lim_{t \to \infty}\frac1{20}\ln\frac{(t+3)^2}{1+t^2}=\frac 1 {20}\ln1=0$$ $$=0+\frac{3 \pi}{20}-\left(\frac{1}{20}\ln9+0\right)$$

$$=\frac{3\pi-\ln9}{20}$$

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    $\begingroup$ But still while substituting $\infty$ , Doesn't the answer become $\log \infty$ which is $\infty$ , so how did you get finite term ? $\endgroup$ – user589548 Oct 4 '18 at 1:55
  • $\begingroup$ no it doesn't see the updated answer $\endgroup$ – Deepesh Meena Oct 4 '18 at 4:51
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    $\begingroup$ And suppose we don't combine $\frac{1}{10} \log {3+t}$ and $\frac{1}{20} \log {1+t^2}$ , Then won't we be able to substitute $\infty$ ??? $\endgroup$ – user589548 Oct 4 '18 at 9:45
  • $\begingroup$ then you will obtain $\infty-\infty$ form and do you know how do you solve it? $\endgroup$ – Deepesh Meena Oct 4 '18 at 11:20
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    $\begingroup$ Noo and that’s why I wrote whether I am right in my solution. $\endgroup$ – user589548 Oct 4 '18 at 11:21

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