Question

${\int }_{-2\pi }^{2\pi }\frac{si{n}^{6}x}{(si{n}^{6}x+co{s}^{6}x)(1+e^{-x})}dx=$

• A. $2\pi$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $4\pi$

Answer 1

The correct option is B $\pi$

The given problem can be easily solved if we know the properties of definite integration.
${\int }_{-a}^{a}f\left(x\right)dx={\int }_0^a\left(f\right(x)+f(-x\left)\right)dx$
Using this, we get -
${\int }_0^{2\pi }\frac{si{n}^{6}x}{(si{n}^{6}x+co{s}^{6}x)(1+e^{-x})}+\frac{si{n}^{6}x}{(si{n}^{6}x+co{s}^{6}x)(1+e^x)}dx$
= ${\int }_0^{2\pi }\frac{si{n}^{6}x}{(si{n}^{6}x+co{s}^{6}x)}dx$
We also know another property which is -
${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$
if f(2a-x) = f(x)
Using this property we get -
= $2{\int }_0^{\pi }\frac{si{n}^{6}x}{(si{n}^{6}x+co{s}^{6}x)}dx$
Again using this property again we can write -
$4{\int }_0^{\frac{\pi }{2}}\frac{si{n}^{6}x}{(si{n}^{6}x+co{s}^{6}x)}dx$
We know an another property which is -
${\int }_a^{b}f\left(x\right)dx={\int }_0^{a}f(a+b-x)dx$
Let I = ${\int }_0^{\frac{\pi }{2}}\frac{si{n}^{6}x}{(si{n}^{6}x+co{s}^{6}x)}dx$
then I = ${\int }_0^{\frac{\pi }{2}}\frac{co{s}^{6}x}{(co{s}^{6}x+si{n}^{6}x)}dx$
Adding both we get -
2I = ${\int }_0^{\frac{\pi }{2}}1.dx$
2I = $\frac{\pi }{2}$
I = $\frac{\pi }{4}$
So, the value of the integral which is asked to find will be = 4.I = $\pi$