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A challenging Integral

sbhatnagar

Active member
Jan 27, 2012
95
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
Jan 28, 2012
150
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.
 
Last edited:

sbhatnagar

Active member
Jan 27, 2012
95
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 \( u=1+\sin(2x) \),

\[ -\int \frac{1}{u\sqrt{u^2-1}}du =-\sec^{-1}(u)+C=-\sec^{-1}(\sin(2x)+1)+C \]
 
Last edited:

oasi

New member
Mar 14, 2012
14
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
Jan 27, 2012
95

Krizalid

Active member
Feb 9, 2012
118
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.