How to evaluate $\int \frac {dx}{\cos x+C}$?

Question

I want to evaluate -

$$\int \frac {dx}{\cos x+C}$$

Where $C$ is an arbitrary constant.I tried substitution and parts but could not do it.

Note that for $C=1$ one can simply do this by using compound angle formulas.But what about other values of $C $?

Thanks for any help!!

score 3 · Answer 1 · answered Jun 12 '17 at 13:50

3

Hint: use the substitution $t=\tan{\frac{1}{2}x}$. Then $dx = 2dt/(1+t^2)$, $$ \cos{x} = \frac{1-t^2}{1+t^2}, $$ and the integral reduces to that of a fairly simple rational function.

answered Jun 12 '17 at 13:50

Chappers

67,606

This is known as Weierstrass substitution – projectilemotion Jun 12 '17 at 13:53
@projectilemotion Thanks a lot!!:-) – Soham Jun 12 '17 at 13:55
1

Although Euler was apparently first to use it in integrals: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution#Euler_and_Weierstrass – Chappers Jun 12 '17 at 13:55

score 2 · Answer 2 · 2017-06-12T14:00:03.787

2

Hint: Make the substitution $ \theta = \frac{x}{2}$.

\begin{align} \frac{1}{C + \cos 2 \theta} &= \frac{1}{C -1 + 2\cos ^2 \theta} \\ &= \frac{\sec^2 \theta}{(C-1)\sec^2 \theta + 2} \\ &= \frac{\sec^2 \theta}{C + 1 + (C-1)\tan^2 \theta} \end{align} Hence

\begin{align} \int \frac{1}{C + \cos x} \text{d}x &= \int \frac{2}{C + \cos 2 \theta} d \theta \\ &= \int \frac{ 2\sec^2 \theta}{ C + 1 + (C-1)\tan^2 \theta} d \theta \end{align}

Now make the second substitution $t = \tan \theta$ and continue as above.

edited Jun 12 '17 at 14:00

answered Jun 12 '17 at 13:54

Typo there... it should be theta instead of y – Soham Jun 12 '17 at 13:56
1

@tatan Thanks! Wrote one variable on my notebook and another when it came to my answer. – Jun 12 '17 at 14:00

How to evaluate $\int \frac {dx}{\cos x+C}$?

2 Answers2