Two different answers for the same integral?

Hijaz Aslam
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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?
 
Physics news on Phys.org
$$\tan^{-1}\left( \frac{1}{x} \right) = \left\{
\begin{array}{lr}
\frac{\pi}{2} - \tan^{-1}(x) : x>0\\
-\frac{\pi}{2} - \tan^{-1}(x) : x<0
\end{array}
\right.
$$
 
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Likes Hijaz Aslam
Oh yes! That cleared it out. Thanks blue_leaf77.
 
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