How to solve $\int_{0}^{\frac{π}{2}} \frac{dx}{4\sin^2(x) +5\cos^2(x)} $ $?$ I apply the substitutions:
$$t=\tan(x), \sin(x)=\frac{t}{\sqrt{1+t^2}}, \cos(x)=\frac{1}{\sqrt{1+t^2}}\ \&\ dx=\frac{dt}{1+t^2}$$
(using $t=\tan(x)$ you can draw a right angled triangle to find the other substitutions)
So we get:
$$\int_{0}^{\frac{π}{2}} \frac{1}{\frac{4t^2}{1+t^2}+\frac{5}{1+t^2}}\frac{dt}{1+t^2}=\frac{1}{4}\int_{0}^{\frac{π}{2}} \frac{1}{t^2+\Big(\frac{\sqrt{5}}{2}\Big)^2} dt$$
This is in a standard integral form, thus:
$$\frac{1}{4}\cdot\frac{2}{\sqrt{5}}\tan^{-1}\Bigg(\frac{2t}{\sqrt{5}}\Bigg)=\frac{1}{2\sqrt{5}}\tan^{-1}\Bigg(\frac{2\tan(x)}{\sqrt{5}}\Bigg)$$
this is from $0$ to $π/2$.
But I can't substitute for $π/2$, because $\tan(x)$ is undefined for $π/2$. If I input this integral into Mathcad I get $\frac{π\sqrt{5}}{20}$. How can I get the right answer out of this? Thanks in advance!
 A: By replacing $x$ with $\arctan t$,
$$ I = \int_{0}^{\pi/2}\frac{dx}{4\sin^2 x+5\cos^2 x} = \int_{0}^{+\infty}\frac{dt}{4t^2+5}=\color{red}{\frac{\pi}{4\sqrt{5}}}.$$
A: Let $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{1}{4\sin^2 x+5\cos^2 x}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^2 x\;,$ We get
$$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{4\tan^2 x+5}dx\;,$$
Now Put $\displaystyle 2\tan x=t\;,$ Then $\displaystyle 2\sec^2 xdx =dt\Rightarrow \sec^2 xdx = \frac{1}{2}dt$
and Changing Limit , We get
$$\displaystyle I = \frac{1}{2}\int_{0}^{\infty}\frac{1}{t^2+5}dt = \frac{1}{2\sqrt{5}}\left[\tan^{-1}\left(\frac{t}{\sqrt{5}}\right)\right]_{0}^{\infty}=\frac{1}{2\sqrt{5}}\cdot \frac{\pi}{2}=\frac{\pi}{4\sqrt{5}}.$$
A: Notice, $$\int_{0}^{\pi/2}\frac{dx}{4\sin^2x+5\cos^2 x}$$
$$=\int_{0}^{\pi/2}\frac{dx}{\cos^2 x\left(4\frac{\sin^2x}{\cos^2 x}+5\right)}$$
$$=\int_{0}^{\pi/2}\frac{\sec^2 x\ dx}{5+4\tan^2 x}$$
$$=\frac{1}{4}\int_{0}^{\pi/2}\frac{d(\tan x)}{\left(\frac{\sqrt 5}{2}\right)^2+(\tan x)^2}$$
$$=\frac{1}{4}\frac{2}{\sqrt 5}\left[\tan^{-1}\left(\frac{2\tan x}{\sqrt 5}\right)\right]_{0}^{\pi/2}$$
$$=\frac{1}{2\sqrt 5}\left[\frac{\pi}{2}-0\right]=\color{blue}{\frac{\pi}{4\sqrt 5}}$$
A: Suppose we seek to evaluate
$$\int_0^{\pi/2} \frac{1}{4\sin^2(x)+5\cos^2(x)} dx
= \frac{1}{4} \int_0^{2\pi} \frac{1}{4\sin^2(x)+5\cos^2(x)} dx.$$
Introduce $z=\exp(ix)$ so that $dz=iz \; dx$ to get
$$\frac{1}{4}\int_{|z|=1}
\frac{1}{4(z-1/z)^2/(-4)+5(z+1/z)^2/4} \frac{dz}{iz}
\\ = \int_{|z|=1}
\frac{1}{-4(z-1/z)^2+5(z+1/z)^2} \frac{dz}{iz}
\\ = \int_{|z|=1}
\frac{z^2}{-4(z^2-1)^2+5(z^2+1)^2} \frac{dz}{iz}
\\ = \frac{1}{i} \int_{|z|=1}
\frac{z}{-4(z^2-1)^2+5(z^2+1)^2} \; dz
\\ = \frac{1}{i} \int_{|z|=1}
\frac{z}{z^4 + 18z^2 + 1} \; dz.$$
We have $$z^4 + 18z^2 + 1 = (z^2+9)^2 - 80$$
so we get for the poles
$$\rho_{1,2,3,4} = \pm\sqrt{\pm\sqrt{80} -9}.$$
Now have by inspection that only the two poles
$$\rho_{1,2} = \pm\sqrt{\sqrt{80} -9}$$
contribute being inside the unit circle and we get for the integral
$$\frac{1}{i} 2\pi i
(\mathrm{Res}(f(z); z=\rho_1)
+ \mathrm{Res}(f(z); z=\rho_2))$$
where we have set
$$f(z) = \frac{z}{z^4 + 18z^2 + 1}.$$
These residues are
$$\left.\frac{z}{4z^3+36z}\right|_{z=\rho_{1,2}}
= \left.\frac{1}{4z^2+36}\right|_{z=\rho_{1,2}}
= \frac{1}{4}\left.\frac{1}{z^2+9}\right|_{z=\rho_{1,2}}.$$
By definition of $\rho_{1,2}$ we have $\rho_{1,2}^2 + 9
= \sqrt{80}$ so we finally obtain
$$\frac{1}{i} 2\pi i \times 2\times\frac{1}{4} \frac{1}{\sqrt{80}}
= \pi \frac{1}{\sqrt{80}}
= \frac{\pi}{4\sqrt{5}}.$$
A: Hint: $4\sin^2x+5\cos^2x=4+\cos^2x$
A: The integral is (almost) the polar form of the ellipse with major radius $\sqrt{5}$ and minor radius $2$. Multiply by $\frac{20}{20}$ to ensure the essential $(ab)^2$ is present in the numerator, and move out a factor of two to put it in the form $\frac{r^2}{2}$, then the integral is simply one tenth of one quarter of the area of the ellipse with the aforementioned radii.
Thus, the integral is equal to $\frac{ \pi \times 2 \times \sqrt{5}}{10 \times 4} = \frac{\pi}{4 \sqrt{5}}$
A: $$
\begin{aligned}
 \int_{0}^{\frac{\pi}{2}} \frac{d x}{4 \sin ^{2} x+5 \cos ^{2} x} =& \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{4 \tan ^{2} x+5} d x \\
=& \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{d(2 \tan x)}{(2 \tan x)^{2}+(\sqrt{5})^{2}} \\
=& \frac{1}{2 \sqrt{5}}\left[\tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)\right]_{0}^{\frac{\pi}{2}} \\
=& \frac{1}{2 \sqrt{5}} \cdot \frac{\pi}{2} \\
=& \frac{\pi}{4 \sqrt{5}}
\end{aligned}
$$
