# A challenging Integral

#### sbhatnagar

##### Active member
Evaluate the integral

$\int \frac{\sin(x)-\cos(x)}{(\sin{(x)}+\cos{(x)})\sqrt{\sin(x)\cos(x)+ \sin^2(x)\cos^2(x)}} dx$

The problem above is not necessarily difficult; however, it can be almost impossible to evaluate if one doesn’t know the right “trick”.

#### Siron

##### Active member
I would do the following:
Let $\sin(x)+\cos(x)=t \Rightarrow [\cos(x)-\sin(x)]dx=dt \Rightarrow -[(\sin(x)-\cos(x)]dx=dt \Rightarrow [\sin(x)-\cos(x)]dx=-dt$
and $\sin(x)\cos(x)=\frac{t^2-1}{2}$

Thus, the integral becomes:
$- \int \frac{dt}{t\sqrt{\frac{t^2-1}{2}\left(1+\frac{t^2-1}{2}\right)}}=-2 \int \frac{dt}{t\sqrt{t^4-1}}=\frac{-1}{2} \int \frac{4t^3}{t^4\sqrt{t^4-1}}$

Let $t^4-1= u \Rightarrow 4t^3dt=du$ so the integral becomes:
$\frac{-1}{2} \int \frac{du}{(u+1)\sqrt{u}}=-\arctan(\sqrt{u})$

Doing the back-substitution we obtain:
$- \arctan\left(\sqrt{[\sin(x)+\cos(x)]^4-1}\right)+C$

I'm not sure my attempt is correct.

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#### sbhatnagar

##### Active member
Hi Siron! You made it. Here's my idea:

\begin{align*} \int \frac{\sin(x)-\cos(x)}{(\sin{(x)}+\cos{(x)})\sqrt{\sin(x)\cos(x)+ \sin^2(x)\cos^2(x)}} dx &= -\int \frac{\cos^2(x)-\sin^2(x)}{(1+2\sin{(x)}\cos{(x)})\sqrt{\sin(x) \cos(x)(\sin(x)\cos(x)+1)}} dx\\ &= -\int \frac{\cos(2x)}{(1+\sin(2x))\sqrt{\frac{\sin(2x)}{2} \left( \frac{\sin(2x)}{2}+1 \right)}} dx \\ &= -\int \frac{2\cos(2x)}{(1+\sin(2x))\sqrt{\sin(2x)(\sin(2x)+2)}} dx\end{align*}

By the substitution $$u=1+\sin(2x)$$,

$-\int \frac{1}{u\sqrt{u^2-1}}du =-\sec^{-1}(u)+C=-\sec^{-1}(\sin(2x)+1)+C$

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#### oasi

##### New member
Hi Siron! You made it. Here's my idea:

\begin{align*} \int \frac{\sin(x)-\cos(x)}{(\sin{(x)}+\cos{(x)})\sqrt{\sin(x)\cos(x)+ \sin^2(x)\cos^2(x)}} dx &= -\int \frac{\cos^2(x)-\sin^2(x)}{(1+2\sin{(x)}\cos{(x)})\sqrt{\sin(x) \cos(x)(\sin(x)\cos(x)+1)}} dx\\ &= -\int \frac{\cos(2x)}{(1+\sin(2x))\sqrt{\frac{\sin(2x)}{2} \left( \frac{\sin(2x)}{2}+1 \right)}} dx \\ &= -\int \frac{2\cos(2x)}{(1+\sin(2x))\sqrt{\sin(2x)(\sin(2x)+2)}} dx\end{align*}

By the substitution $$u=1+\sin(2x)$$,

$-\int \frac{1}{u\sqrt{u^2-1}}du =-\sec^{-1}(u)+C=-\sec^{-1}(\sin(2x)+1)+C$
why we chose u=1+\sin(2x)

#### sbhatnagar

##### Active member
Why substitute $u=1+\sin(2x)$?
...because it make the solution easy.

#### Krizalid

##### Active member
I wouldn't consider that argument enough to say why it works, and actually, the answer is very simple, for the one who asked why it works, just check the integrand, and see the derivative of the substitution involved, everything works nicely.