Question

The integral $\int_{0}^{\pi/4} \frac{cosx-sinx}{2+sin(2x)} dx$ equals

Answer 1

$\arctan(\sqrt{2}) - \frac{\pi}{4}$

Answer 2

The integral to evaluate is $I = \int_{0}^{\pi/4} \frac{cosx-sinx}{2+sin(2x)} dx$.

We observe the numerator $cosx-sinx$. This is the derivative of $sinx+cosx$. Let's make the substitution $t = sinx+cosx$. Then, differentiating $t$ with respect to $x$, we get $dt = (cosx-sinx)dx$.

Next, we need to express the term $sin(2x)$ in the denominator in terms of $t$. We know that $t^2 = (sinx+cosx)^2 = sin^2x+cos^2x+2sinxcosx$. Since $sin^2x+cos^2x=1$ and $2sinxcosx=sin(2x)$, we have: $t^2 = 1+sin(2x)$ From this, we can write $sin(2x) = t^2-1$.

Now, we need to change the limits of integration according to the substitution: When $x=0$: $t = sin(0)+cos(0) = 0+1 = 1$. When $x=\pi/4$: $t = sin(\pi/4)+cos(\pi/4) = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Substitute these into the integral: $I = \int_{1}^{\sqrt{2}} \frac{dt}{2+(t^2-1)}$ $I = \int_{1}^{\sqrt{2}} \frac{dt}{t^2+1}$ This is a standard integral of the form $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$. Here, $a=1$. $I = \left[\arctan(t)\right]_{1}^{\sqrt{2}}$ Now, apply the limits of integration: $I = \arctan(\sqrt{2}) - \arctan(1)$ We know that $\arctan(1) = \frac{\pi}{4}$. Therefore, the value of the integral is: $I = \arctan(\sqrt{2}) - \frac{\pi}{4}$