# Why does integral $\int \frac 1{(x+\cos a)^2+(\sin a)^2} dx$ have two solutions

For the integral $$\int \frac 1{(x+\cos a)^2+(\sin a)^2} dx$$

My evaluation

$$\csc a\arctan\left(\frac {x+\cos a}{\sin a}\right)$$

Wolfram Alpha's evaluation here

$$\csc a\arctan\left(\frac {(x-1)\tan\frac a2}{x+1}\right)$$

When I was doing a STEP question I found this integral and evaluated it. I checked it with Wolfram Alpha and it gave me a different solution, so I checked on another integration calculator that gave my answer. In fact, Wolfram Alpha was the only website to give their evaluation of the integral.

This intrigued me so i graphed both Mine and Wolfram Alpha's equation on Desmos; the graphs were different but both valid solutions to the same integral.

If anyone knows why this happened or how to get Wolfram Alpha's solution please let me know. Also if there are any other examples of integrals you have come across that have multiple unique solutions please share.

• An antiderivative is a function $F$ that satisfies $F'=f$. Antiderivative refers to a family of curves that satisfy the above relation. They all differ from each other by a constant value due to the process of differentiation. Your evaluations look different but Quanto's answer explains that they only differ by a constant. Nov 12, 2019 at 19:10
• If you read carefully you'll see that Wolfram Alpha's solution is $$\csc a\arctan\left(\frac {(x-1)\tan\frac a2}{x+1}\right) \color{red}{+\text{constant}}$$, which should remind you that there is an infinity of solutions (not only two). Jun 17 at 15:29

The two answers differs by a constant, as shown below. Let $$u= \frac {x-1}{x+1}\tan\frac a2$$, $$v=\frac {\sin a}{x+\cos a}$$ and evaluate

$$\frac{ u + v}{1-uv}= \tan\frac a2 \cdot \frac{(x+\cos a)(x-1) + 2\cos^2 \frac a2 (x+1)}{(x+\cos a)(x+1)-2\sin^2 \frac a2 (x-1)}=\tan\frac a2$$

Then, utilize the identity $$\arctan \frac{ u + v}{1-uv} =\arctan u + \arctan v$$ to get

$$\arctan\left(\frac {(x-1)\tan\frac a2}{x+1}\right)+\arctan\left(\frac {\sin a}{x+\cos a}\right)=\arctan\left(\tan\frac a2\right) = \frac a2$$

Also, note that

$$\arctan\left(\frac {\sin a}{x+\cos a}\right) = \frac\pi2 - \arctan\left(\frac {x+\cos a}{\sin a}\right)$$

Thus, the difference of the two is

$$\csc a\arctan\left(\frac {(x-1)\tan\frac a2}{x+1}\right) -\csc a\arctan\left(\frac {x+\cos a}{\sin a}\right) = \frac12(a-\pi)\csc a$$

which is a constant, as expected.