Evaluating $ \int \frac{1}{5 + 3 \sin(x)} ~ \mathrm{d}{x} $.

Question

What is the integral of: $\int \frac{1}{5+3\sin x}dx$

My attempt:

Using: $\tan \frac x 2=t$, $\sin x = \frac {2t}{1+t^2}$, $dx=\frac {2dt}{1+t^2}$ we have:

$\int \frac{1}{5+3\sin x}dx= 2\int \frac 1 {5t^2+6t+5}dt $

I'll expand the denominator: $5t^2+6t+5=5((t+\frac 3 5 )^2+1-\frac 1 4 \cdot (\frac 6 5)^2)=5((t+\frac 3 5)^2+0.64)$. So:

$2\int \frac 1 {5t^2+6t+5}dt = \frac 2 5 \int \frac 1{(t+\frac 3 5)^2+0.64}dt=\frac 2 5(\frac 5 4\arctan((t+\frac 3 5)\frac 5 4))=\frac 1 2 \arctan(\frac{5t+3}{4}) $

But if I'll place $\tan \frac x 2=t$ I won't be able to simplify it further and since the online calculator's answers don't have $\tan \frac x 2$ there, I believe I made a mistake. What is wrong with what I did and is there a better way to do it?

Answer 1

Hint:

From what you got on that calculator, use the following identities to simplify it:

• $\sin(x)=2\sin(x/2)\cos(x/2)$
• $1+\cos(x)=2\cos^2(x/2)$

You'll see that the results are same.

Answer 2

Its very likely that the expression was simplified further from $\tan(x/2)$ as that expression has many identities, hence why its revered for its integral substitution properties. Also, your answer is correct. Your method is very brute force, but if you know what your doing, your not overkilling the problem that much.

Answer 3

Taking from $5t^2$+6t+5, instead of expanding, complete the square.

=5(t+$\frac{6}{10})^2$-$\frac{16}{5}$

Rewrite in the form

=$\frac{-1}{5}$$\int$$\frac{1}{16/25-(t+6/10)^2}$

=$\frac{-1}{5}$[$\frac{5}{4}$$tanh^{-1}$($\frac{t+6/10}{4/5})]$+C

=$\frac{-1}{4}$[$tanh^{-1}$($\frac{5t}{4}$+$\frac{3}{2}$)]+C

Still getting answer in terms of tan(x/2).

Answer 4

Here is a cleaner evaluation \begin{align} & \int \frac{1}{5 + 3 \sin x}dx\\ =& \int \frac{\sec x}{5 \sec x+ 3 \tan x}dx = \int \frac{5 \sec^2 x+ 3 \sec x\tan x}{(5 \sec x+ 3 \tan x)^2}dx\\ = &\int \frac{d (5 \tan x+ 3 \sec x)}{(5 \tan x+ 3 \sec x)^2+16}=\frac14\tan^{-1}\frac{5 \tan x+ 3 \sec x}4 \end{align}