Trigonometric Integral : $\int\frac{1}{\sin x+ 3\cos x}dx$

Question

I would appreciate if somebody could help me with the following problem

Q: How to integrate this integral $$\int\frac{1}{\sin x+ 3\cos x}dx$$

Perhaps you could use a more informative title? – davidlowryduda Dec 10 '13 at 01:44 — davidlowryduda, Dec 10 '13 at 01:44
Perhaps by using the tangent half-angle substitution ? – Lucian Dec 10 '13 at 01:54 — Lucian, Dec 10 '13 at 01:54
@Lucian that's the answer to every question :) – Igor Rivin Dec 10 '13 at 02:09 — Igor Rivin, Dec 10 '13 at 02:09
@IgorRivin: Even to this one ? :-) – Lucian Dec 10 '13 at 02:42 — Lucian, Dec 10 '13 at 02:42
@Lucian especially this one... – Igor Rivin Dec 10 '13 at 02:58 — Igor Rivin, Dec 10 '13 at 02:58

score 4 · Accepted Answer · answered Dec 10 '13 at 02:06

\begin{align*} \int \frac 1 {\sin x +3\cos x}\ dx &= \int \frac 1 {2\sin \frac x 2\cos \frac x 2 +3\cos^2 \frac x 2 - 3 \sin^2 \frac x 2}\ dx \\ &= \int \frac 1 {\cos^2 \frac x 2} \cdot \frac 1 {3 + 2\tan \frac x 2 - 3 \tan^2 \frac x 2}\ dx \end{align*}

Switching $y=\tan \frac x 2 \implies\frac {dy}{dx} = \frac 1 2 \frac 1 {\cos^2 \frac x 2}$

\begin{align*} \int \frac 1 {\sin x +3\cos x}\ dx &= \int \frac 1 {3 + 2y - 3 y^2}2\ dy \\ &= 6\int \frac 1 {(3y-1-\sqrt{10})(3y-1+\sqrt{10})} \ dy \\ &= \frac 3 {\sqrt{10}}\int \frac 1 {3y-1-\sqrt{10}} - \frac 1 {3y-1+\sqrt{10}} \ dy \\ \end{align*}

I'll leave the rest.

score 4 · Answer 2 · answered Dec 10 '13 at 02:12

4

$$1\cdot \sin x +3 \cdot \cos x=\sqrt{10} \cdot \sin (x+a)$$ where $a=\tan ^{-1}(3)$

So, now can you just apply the known integral of $\csc(x+a) $?

answered Dec 10 '13 at 02:12

DJohnM

3,630

Trigonometric Integral : $\int\frac{1}{\sin x+ 3\cos x}dx$

2 Answers2