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I'm asked to evaluate for $a,b \gt 0$ the integral $$\int^{\pi/2}_0 \frac{1}{a\text{sin}^2x+b\text{cos}^2x}dx.$$

Here's how I go with it:

I factorise the denominator:

$a\text{sin}^2x+b\text{cos}^2x=\text{sin}^2\Bigl(a+\frac{b\text{cos}^2x}{\text{sin}^2x}\Bigr)=\text{sin}^2x(a+b\text{cotg}^2x)$

Hence, the integral $$\int^{\pi/2}_0 \frac{1}{a\text{sin}^2x+b\text{cos}^2x}dx=\int^{\pi/2}_0\frac{1}{\text{sin}^2x(a+b\text{cotg}^2x)}dx$$

I make a substitution: $u=\text{cotg}x$ $\Rightarrow du=-\frac{1}{\text{sin}^2x}dx$

$\Rightarrow$ $$-\int^{u_{\pi/2}}_{u_0} \frac{1}{a+bu^2}du$$I make a substitution:

$$t=\frac{\sqrt{b}u}{\sqrt{a}} \Rightarrow dt=\frac{\sqrt{b}}{\sqrt{a}}du$$

$\Rightarrow$ $$-\int^{t_{\pi/2}}_{t_0} \frac{\sqrt{a}}{\sqrt{b}(at^2+a)}dt$$

$$=-\frac{\sqrt{a}}{\sqrt{b}a}\int^{t_{\pi/2}}_{t_0}\frac{1}{t^2+1}=$$

$$\frac{1}{\sqrt{ab}}\text{arctan}(t)\Bigg|_{t_0}^{t_{\pi/2}}=$$

$$\frac{1}{\sqrt{ab}}\text{arctan}\left(\frac{\sqrt{b}u}{\sqrt{a}}\right)\Bigg|_{u_0}^{u_{\pi/2}}=$$

$$\frac{1}{\sqrt{ab}}\text{arctan}\left(\frac{\sqrt{b}}{\sqrt{a}}\text{cotg}x\right)\Bigg|^{\pi/2}_0=$$

$$\frac{1}{\sqrt{ab}}\Biggl(\text{arctan}(0)-\text{undefined cotg value}\Biggr)$$

I'm getting an undefined value when I evaluate the definite integral since $\text{cotg}(0)$ is undefined?

Am I doing something wrong?

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  • $\begingroup$ The problem is you have not changed the limits for your new variable. For another approach, note that $\sin^2\theta=\dfrac{2t}{1+t^2}$ and $\cos^2\theta=\dfrac{1-t^2}{1+t^2},$ where $t=\tan\theta.$ Then use a substitution accordingly :) Also, the usual convention to write trig-functions is to use three letters. $\endgroup$ – Bumblebee Mar 23 at 17:27
  • $\begingroup$ You forgot to change the bounds, after doing your first substitution, you'll get $ \int\limits_{0}^{+\infty}{\frac{\mathrm{d}u}{a+bu^{2}}} $, after doing the second one, you'll get $ \frac{1}{\sqrt{ab}}\int\limits_{0}^{+\infty}{\frac{\mathrm{d}t}{1+t^{2}}}=\frac{\pi}{2\sqrt{ab}} $, assuming $ a $ and $ b $ are different from $ 0 \cdot $ $\endgroup$ – CHAMSI Mar 23 at 17:35
  • $\begingroup$ You did not change the bounds. $\endgroup$ – hamam_Abdallah Mar 23 at 17:39
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By the substitution $ u=cotg(x)$, the integral becomes

$$-\int_{+\infty}^0\frac{du}{a+bu^2}$$

$$=\frac{1}{a}\int_0^{+\infty}\frac{du}{1+\frac ba u^2}$$

now, put $v=u\sqrt{\frac ba}$

it gives $$\frac{1}{a}\int_0^{+\infty}\frac{\sqrt {\frac ab}dv}{1+v^2}$$

$$=\frac{1}{\sqrt{ab}}\Bigl[\arctan(v)\Bigr]_0^{+\infty}=\frac{\pi}{2\sqrt{ab}}$$

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Hint

Another way

$$\dfrac1{a\sin^2x+b\cos^2x}=\dfrac{\sec^2x}{a\tan^2x+b}$$

Let $\sqrt a\tan x=\sqrt b\tan y$

$x=0\implies y=0$ and $x=\dfrac\pi2\implies y=?$

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