Integration $\int_0^{\pi/2} \frac{dx}{(3 + 5 \cos x)^2}$

Question

I had tried to solve this integral; using the substitution $\tan(x/2) =t$, and $\cos x= \frac{1-t^2}{1+t^2}$. But after making terms in $t$, I am not able to integrate further as numerator contains quadratic and denominator contains biquadratic.

$\int\limits_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}\ dx$.

Answer 1

Upon making the substitution, the integral should become

\begin{equation*} \int_{0}^{\pi/2}{\frac{1}{\left[3+5\cos{(x)}\right]^{2}}\,\mathrm{d}x} = \int_{0}^{1}{\frac{t^{2}+1}{2t^{4} - 16t^{2}+32}\,\mathrm{d}t}. \end{equation*}

From here, as mentioned in a comment, you should be able to use a partial fraction decomposition to evaluate this integral.

Answer 2

Well let's use Weierstrass and partial fractions ig and evaluate the definite with FTC II.

Using Weierstrass with the tangent half angle substitution, our integral becomes $$\int \frac{1}{(3 + 5 \cos x)^2}\ dx = \int \frac{1}{\left(3+5\left(\frac{1-t^2}{1+t^2}\right)\right)^2}\cdot \frac{2dt}{1+t^2} = \frac12\int {t^2 + 1\over(t - 2)^2 (t + 2)^2}dt$$

Perform partial fraction decomposition by setting it up like this $$ {t^2 + 1\over2(t - 2)^2 (t + 2)^2}dt = \frac{A}{2(t-2)} + \frac{B}{2(t-2)^2} + \frac{C}{2(t+2)} + \frac{D}{2(t+2)^2}$$ Then just multiply by the denominator, match powers, and solve the system of equations for the unknown. We get

$$ {t^2 + 1\over2(t - 2)^2 (t + 2)^2}dt = \frac{3}{64(t-2)} + \frac{5}{32(t-2)^2} + \frac{-3}{64(t+2)} + \frac{5}{32(t+2)^2}$$

Now integrate termwise (log for single power, 1/(whatever) for double power) and simplify to get $$\int \frac{3}{64(t-2)} + \frac{5}{32(t-2)^2} + \frac{-3}{64(t+2)} + \frac{5}{32(t+2)^2} dt $$$$= {-5t\over 16(t^2-4)} + {3\over 64}\ln(2-t) -\frac3{64}\ln(t+2)+C$$

The bounds transform as follows $\tan\left({\frac\pi2\over2}\right)=1$ and the lower one remains $0$ so $${-5t\over 16(t^2-4)} + {3\over 64}\ln(2-t) -\frac3{64}\ln(t+2)\Big|^1_0 = \frac5{48} - {3\ln(3)\over64}$$

Answer 3

Let's take $\int_0^{\pi/2} \frac{1}{\left(3+5 \cos\left(x\right)\right)^2}\text{d}x$ and substitute $u = \tan\left(\frac{x}{2}\right)$ and $\text{d}u = \frac{1}{2}\text{d}x\sec^2\left(\frac{x}{2}\right)$ which yields $\cos\left(x\right) = \frac{1 - u^2}{1 + u^2}$ and $\text{d}x = \frac{2\text{d}u}{1 + u^2}$:

$$ \begin{align} \int_0^{\pi/2} \frac{1}{\left(3+5 \cos\left(x\right)\right)^2}\text{d}x &= \int_{\tan\left(0\right)}^{\tan\left(\pi/4\right)} \frac{2}{\left(1 + u^2\right)\left(\frac{5\left(1-u^2\right)}{1+u^2}+3\right)^2} \text{d}u \\ &= \int_0^1 \frac{u^2+1}{2\left(u^2-4\right)^2} \text{d}u \end{align} $$ Now decompose $\frac{u^2+1}{2\left(u^2-4\right)^2} = \frac{u^2+1}{2\left(u-2\right)^2\left(u+2\right)^2}$ into its partial fractions: $$ \begin{align} &\frac{u^2+1}{2\left(u-2\right)^2\left(u+2\right)^2} &&= \frac{\theta_1}{2\left(u-2\right)} + \frac{\theta_2}{2\left(u-2\right)^2} + \frac{\theta_3}{2\left(u+2\right)} + \frac{\theta_4}{2\left(u+2\right)^2} \\ \Leftrightarrow &\; u^2 + 1 &&= \theta_1\left(u+2\right)^2\left(u-2\right)+\theta_2\left(u+2\right)^2+\theta_3\left(u+2\right)\left(u-2\right)^2\\&&&\quad+\theta_4\left(u-2\right)^2 \\ &&&=\left(\theta_1+\theta_3\right)u^3+\left(2\theta_1+\theta_2-2\theta_3+\theta_4\right)u^2 \\&&&\quad+\left(-4\theta_1+4\theta_2-4\theta_3-4\theta_4\right)u+\left(-8\theta_1+4\theta_2+8\theta_3+4\theta_4\right) \end{align} $$ This gives us $4$ equations in $4$ unknowns: $$ \begin{align} 1&=-8\theta_1+4\theta_2+8\theta_3+4\theta_4 \\ 0&=-4\theta_1+4\theta_2-4\theta_3-4\theta_4 \\ 1&=2\theta_1+\theta_2-2\theta_3+\theta_4 \\ 0&=\theta_1+\theta_3 \end{align} $$ This system of linear equations can be solved using Gauss-Jordan elimination: $$ \left(\begin{array}{cccc|c} -8 & 4 & 8 & 4 & 1 \\ -4 & 4 & -4 & -4 & 0 \\ 2 & 1 & -2 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 \end{array}\right) \sim \left(\begin{array}{cccc|c} 1 & 0 & 0 & 0 & \frac{3}{32} \\ 0 & 1 & 0 & 0 & \frac{5}{16} \\ 0 & 0 & 1 & 0 & -\frac{3}{32} \\ 0 & 0 & 0 & 1 & \frac{5}{16} \end{array}\right) $$ So the solution to this system is $\theta = \left(\array{\frac{3}{32} \\ \frac{5}{16} \\ -\frac{3}{32} \\ \frac{5}{16}}\right)$ which gives us: $$ \begin{align} \int_0^1 \frac{u^2+1}{2\left(u^2-4\right)^2} \text{d}u &= \int_0^1 \frac{3}{64\left(u-2\right)} + \frac{5}{32\left(u-2\right)^2} \\&\qquad - \frac{3}{64\left(u+2\right)}+\frac{5}{32\left(u+2\right)^2} \text{d}u \tag{1}\label{1}\\ &=\frac{3}{64}\int_0^1 \frac{1}{u-2}\text{d}u + \frac{5}{32}\int_0^1\frac{1}{\left(u-2\right)^2}\text{d}u \\&\quad-\frac{3}{64}\int_0^1\frac{1}{u+2}\text{d}u + \frac{5}{32}\int_0^1\frac{1}{\left(u+2\right)^2}\text{d}u \tag{2}\label{2}\\ &= \left.\frac{3}{64}\log\left(u-2\right)\right\rvert_0^1 - \left.\frac{5}{32\left(u-2\right)}\right\rvert_0^1 \\&\quad- \left.\frac{3}{64}\log\left(u+2\right)\right\rvert_0^1 - \left.\frac{5}{32\left(u+2\right)}\right\rvert_0^1 \\ &=-\frac{3}{64}\log\left(2\right) + \frac{5}{64} - \frac{3}{64}\log\left(\frac{3}{2}\right) + \frac{5}{192} \\ &= \frac{5}{48} - \frac{3}{64}\log\left(3\right) \\ &\approx 0.0527 \end{align} $$

$\eqref{1}$ Integrate term by term.
$\eqref{2}$ Substitute $v_1 = u - 2, \text{d}v_1 = \text{d}u$ and $v_2 = u + 2, \text{d}v_2 = \text{d}u$.

Answer 4

An alternative approach is to integrate both sides of $$\left(\frac{5\sin x}{3 + 5 \cos x}\right)’ = \frac{16}{(3 + 5 \cos x)^2}+\frac3{3 + 5 \cos x}$$ to simply the integral \begin{align} \int_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}dx = \frac{5}{48} -\frac1{16}\int_0^{\pi/2} \frac3{3 + 5 \cos x}dx \end{align}