Two different answers for the same integral?

[Hijaz Aslam]

Homework Statement

The anti-derivative of ∫$\frac{sinx}{sin^2x+4cos^2x}$ is $\frac{1}{\sqrt{3}}tan^{-1}((\frac{1}{\sqrt{3}})g(x))+C$ then $g(x)$ is equal to :

a. $secx$
b. $tanx$
c. $sinx$
d. $cosx$

Homework Equations

$d(cosx)=-sinx dx$

The Attempt at a Solution

I tried the problem the following way:

\int \frac{sinx}{sin^2x+4cos^2x}\, dx=\int \frac{sinx}{1+3cos^2x}\, dx
Let $t=cosx$. Therefore$-dt=sinx dx$
Therefore I=\int \frac{-dt}{1+3t^2}=-\frac{1}{\sqrt{3}}tan^{-1}(\sqrt{3}cosx)

But my text gives the solution the following way:

\int \frac{sinx}{sin^2x+4cos^2x}\, dx=\int \frac{tanxsecx}{tan^2x+4}\, dx=\int \frac{tanxsecx}{sec^2x+3}\, dx. Let $t=secx$, therefore:

I=\int \frac{dt}{t^2+3}=\frac{1}{\sqrt{3}}tan^{-1}(\frac{secx}{\sqrt{3}})

So
My answer is : I_m=-\frac{1}{\sqrt{3}}tan^{-1}(\sqrt{3}cosx)+C
Answer in my textbook is: I_t=\frac{1}{\sqrt{3}}tan^{-1}(\frac{secx}{\sqrt{3}})+C

And, I can't find any way to convert $I_m$ to $I_t$.

Am I wrong somewhere?

 

[blue_leaf77]

$$\tan^{-1}\left( \frac{1}{x} \right) = \left\{
\begin{array}{lr}
\frac{\pi}{2} - \tan^{-1}(x) : x>0\\
-\frac{\pi}{2} - \tan^{-1}(x) : x<0
\end{array}
\right.
$$

 

[Hijaz Aslam]

Oh yes! That cleared it out. Thanks blue_leaf77.