# Homework Help: Two different answers for the same integral?

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1. Jan 17, 2016

### Hijaz Aslam

1. The problem statement, all variables and given/known data
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. $cosx$

2. Relevant equations
$d(cosx)=-sinx dx$

3. 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?

2. Jan 17, 2016

### blue_leaf77

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

3. Jan 17, 2016

### Hijaz Aslam

Oh yes! That cleared it out. Thanks blue_leaf77.