Question

Consider the integral $I={\int }_0^{10\pi }\frac{cos6x.cos7x.cos8x.cos9x}{1+e^{2si{n}^{3}4x}}dx$
If $I=c.{\int }_0^{\frac{\pi }{4}}cos6x.cos8x.cos2xdx$ then 'c' equals

• A. 5
• B. 10
• C. 20
• D. $\frac{5}{2}$

Answer 1

The correct option is A 5

$I={\int }_0^{10\pi }\frac{cos6x.cos7x.cos8x.cos9x}{1+e^{2si{n}^{3}4x}}dx\dots \left(1\right)I={\int }_0^{10\pi }\frac{cos6x.cos7x.cos8x.cos9x}{1+e^{-2si{n}^{3}4x}}dx\left(\text{using}{\int }_0^{a}f\right(x)={\int }_0^{a}f(a-x\left)\right)I={\int }_0^{10\pi }\frac{e^{2si{n}^{3}4x}(cos6x.cos7x.cos8x.cos9x)}{1+e^{2si{n}^{3}4x}}dx\dots \left(2\right)\text{Add}\left(1\right)+\left(2\right)2I={\int }_0^{10\pi }cos6x.cos7x.cos8x.cos9xdx$
Let $f\left(x\right)=cos6x.cos7x.cos8x.cos9x$

$f\left(x\right)\text{repeats after}\pi \text{interval}2I=5{\int }_0^{2\pi }f\left(s\right)dx;againf(2a-x)=f\left(x\right)2I=10{\int }_0^{\pi }f\left(x\right)dx;againf(\pi -x)=f\left(x\right)2I=20{\int }_0^{\frac{\pi }{2}}f\left(x\right)dx\Rightarrow I=10{\int }_0^{\frac{\pi }{2}}f\left(x\right)dx\therefore I=10{\int }_0^{\frac{\pi }{2}}cos6x.cos7x.cos8x.cos9xdx\dots \left(3\right)I=10{\int }_0^{\frac{\pi }{2}}cos(3\pi -6x)cos(\frac{7\pi }{2}-7x)cos(4x-8x)cos(\frac{9\pi }{2}-9x)dxI=10{\int }_0^{\frac{\pi }{2}}cos6x.sin7x.cos8x.sin9xdx\dots \left(4\right)$
$\left(3\right)+\left(4\right)2I=10{\int }_0^{\frac{\pi }{2}}cos6x.cos8x[cos7x.cox9x+sin7x.sin9x]dx2I=10{\int }_0^{\frac{\pi }{2}}cos6x.cos8x.cos2xdxI=5{\int }_0^{\frac{\pi }{2}}cos6x.cos8x.cos2xdx$