How to compute integral of $\frac{1}{5 - 4\sin(x) + 3\cos(x)}$

Question

So, I need to compute a integral for $$ \frac{1}{5-4 \sin(x) + 3 \cos(x)}. $$ On integral calculator the following transformation is made: $$ -\frac{\sec^2\left(\frac{x}{2}\right)}{2\left(\tan\left(\frac{x}{2}\right)-2\right)^2}. $$

I have no idea how the first expression is transformed into the second one. Could someone help me out?

Answer 1

Hint: Use Tangent half-angle substitution $$\sin x=\dfrac{2t}{1+t^2}~~~,~~~\cos x=\dfrac{1-t^2}{1+t^2}~~~,~~~dx=\dfrac{2}{1+t^2}$$ and then $t-2=u$.

Answer 2

Since $$ \sin x = \frac{2\tan\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)} \qquad \text{and} \qquad \cos x = \frac{1- \tan^2\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)} $$ We have \begin{align*} 5 - 4\sin x + 3\cos x & = 5 - 4 \frac{2\tan\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)} + 3\frac{1- \tan^2\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)} \\ & = \frac{5+5\tan^2\left(\frac{x}{2}\right) - 8\tan\left(\frac{x}{2}\right) + 3 - 3\tan^2\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)} \\ & =\frac{2\left(\tan^2\left(\frac{x}{2}\right) - 4\tan\left(\frac{x}{2}\right) + 4\right)}{1+\tan^2\left(\frac{x}{2}\right)} \\ & = \frac{2\left(\tan\left(\frac{x}{2}\right)-2\right)^2}{\sec^2\left(\frac{x}{2}\right)} \end{align*} So, $$ \frac{1}{5 - 4\sin x + 3\cos x} = \frac{\sec^2\left(\frac{x}{2}\right)}{2\left(\tan\left(\frac{x}{2}\right)-2\right)^2}. $$ Then the numerator is the derivative of $ \tan(x/2)-2$. [ $d\left(\tan\left(\frac{x}{2}\right)-2\right) = \frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx$ ].

So, $$I = \int\frac{{d\left( \tan\left(\frac{x}{2}\right)-2\right)}}{\left( \tan\left(\frac{x}{2}\right)-2\right)^2} = -\frac{1}{\tan\left(\frac{x}{2}\right)-2} +c$$

Answer 3

Hint

$$5+3\cos x-4\sin x=5(1+\cos(x+\arccos\dfrac35))$$

$$\cos2y=2\cos^2y-1$$

Answer 4

A variant hint:

Write $3\cos x-4\sin x$ as $\;5\cos(x+\varphi)$, where \begin{cases}\cos\varphi= \frac35,\\ \sin\varphi= \frac45. \end{cases} Then use the substitution $u=x+\varphi$ to obtain \begin{align}\int\frac{\mathrm d x}{5 - 4\sin(x) + 3\cos(x)}&=\int\frac{\mathrm d u}{5(1+\cos u)}=\int\frac{1-\cos u}{5\sin^2 u}\,\mathrm d u=\int\frac{\mathrm d u}{5(1+\cos u)}\\[1ex]&=\int\frac{1-\cos u}{5\sin^2 u}\,\mathrm d u=\frac15\int\frac{\mathrm d u}{\sin^2 u}-\frac15\int\frac{\cos u}{\sin^2 u}\,\mathrm d u. \end{align}