Simplifying gives : $$I= \int_0^{\frac{\pi }{2}} \frac{cos ^2 x+ sin x cos x}{1+2sinx cos x} dx$$$$I= \int_0^{\frac{\pi }{2}} \frac{sin ^2 x+ sin x cos x}{1+2sinx cos x} dx$$ $$2I= \int_0^{\frac{\pi }{2}} \frac{cos ^2 x+sin ^2 x+ 2sin x cos x}{1+2sinx cos x} dx= \int_0^{\frac{\pi }{2}} dx =\frac{\pi}{2}$$ Hence, $$\boxed{I=\frac{\pi}{4}}$$

chem123 wrote: Simplifying gives : $$I= \int_0^{\frac{\pi }{2}} \frac{cos ^2 x+ sin x cos x}{1+2sinx cos x} dx$$$$I= \int_0^{\frac{\pi }{2}} \frac{sin ^2 x+ sin x cos x}{1+2sinx cos x} dx$$ $$2I= \int_0^{\frac{\pi }{2}} \frac{cos ^2 x+sin ^2 x+ 2sin x cos x}{1+2sinx cos x} dx= \int_0^{\frac{\pi }{2}} dx =\frac{\pi}{2}$$ Hence, $$\boxed{I=\frac{\pi}{4}}$$ How did the second line of your solution come?

Its called KING'S RULE. One can replace $x$ with $a+b-x$ in a definite Integral, where $a,b$ are the lower and upper limits

DeZade2002 wrote: How did the second line of your solution come? Just change $x\to\frac{\pi}2-x$

chem123 wrote: Its called KING'S RULE. One can replace $x$ with $a+b-x$ in a definite Integral, where $a,b$ are the lower and upper limits Then does it mean $\int_a^{b} f(x)dx = \int_a^{b} f(a+b-x)dx$?

$$I= \int_0^{\frac{\pi }{2}} \frac{cos x}{sinx+cos x} dx$$$$\Rightarrow I=\dfrac {\pi}{4}$$

DeZade2002 wrote: chem123 wrote: Its called KING'S RULE. One can replace $x$ with $a+b-x$ in a definite Integral, where $a,b$ are the lower and upper limits Then does it mean $\int_a^{b} f(x)dx = \int_a^{b} f(a+b-x)dx$? Yes, it does, because the graph of $f(a+b-x)$ from $a$ to $b$ is just a horizontally flipped version of the graph of $f(x)$, so their integrals must also be equal as well.

$I = \int^{\pi/2}_0 \frac{\cos^4x + \sin x \cos^3 x + \sin^2x\cos^2x + \sin^3x\cos x}{\sin^4x + \cos^4x + 2\sin x\cos^3x + 2\sin^2x\cos^2x + 2\sin^3x\cos x} \ \mathrm{dx}$ Using King's rule $I = \int^{\pi/2}_0 \frac{\sin^4x + \cos x \sin^3 x + \sin^2x\cos^2x + \sin x\cos^3 x}{\sin^4x + \cos^4x + 2\sin x\cos^3x + 2\sin^2x\cos^2x + 2\sin^3x\cos x} \ \mathrm{dx}$ Hence $2I = \int^{\pi/2}_0 \frac{\sin^4x + \cos^4x + 2\sin x\cos^3x + 2\sin^2x\cos^2x + 2\sin^3x\cos x}{\sin^4x + \cos^4x + 2\ sinx\cos^3x + 2\sin^2x\cos^2x + 2\sin^3x\cos x} \ \mathrm{dx} = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4}$

If you want the definite integral, you can multiply top and bottom by cos(x)-sin(x) and use double angle identities. Simplifies quite well.

NikoIsLife wrote: DeZade2002 wrote: chem123 wrote: Its called KING'S RULE. One can replace $x$ with $a+b-x$ in a definite Integral, where $a,b$ are the lower and upper limits Then does it mean $\int_a^{b} f(x)dx = \int_a^{b} f(a+b-x)dx$? Yes, it does, because the graph of $f(a+b-x)$ from $a$ to $b$ is just a horizontally flipped version of the graph of $f(x)$, so their integrals must also be equal as well. Thanks a lot!

Notice that $$\frac{\cos^4x + \sin x \cos^3 x + \sin^2x\cos^2x + \sin^3 x \cos x}{\sin^4 x + \cos^4 x + 2\sin x\cos^3x + 2\sin^2x\cos^2x + 2\sin^3x\cos x}=\frac{\cos^2 x \left (\cos^2 x+\sin^2 x \right)+\sin x \cos x \left (\cos^2 x+\sin^2 x\right)}{\left (\sin^2 x+\cos^2 x\right)^2+2\sin x\cos x \left (\cos^2 x+\sin^2 x\right)}=\frac{\cos^2 x+\sin x \cos x}{1+2\sin x \cos x}=\frac{1}{1+\tan x}.$$If we let $x=\frac{\pi}2-y$ then, $$I:=\int_0^{\pi/2} \frac{1}{1+\tan x} \ dx =\int_0^{\pi/2} \frac{1}{1+\cot y} \ dy$$so, $$I=\frac 12 \int_0^{\pi/2} \left (\frac{1}{1+\tan x}+\frac{1}{1+\cot x}\right) \ dx=\frac 12 \int_0^{\pi/2} \ dx=\frac{\pi}4.$$

Change f(x) —> f(a + b - x) where a is the lower limit and b is the upper limit of the definite integral and then add them to get 2I = Integral dx from 0 to π/2 = π/2 - 0 = π/2 Therefore, I = π/4