Question

The value of integral $\underset{0}{\overset{10\pi }{\int }}\frac{\cos 6x\cos 7x\cos 8x\cos 9x}{1+e^{{\sin }^{3}4x}}dx$ is equal to

• A. $\frac{\pi }{8}$
• B. $\frac{3\pi }{8}$
• C. $\frac{5\pi }{8}$
• D. $\frac{7\pi }{8}$

Answer 1

The correct option is C $\frac{5\pi }{8}$

$I=5\underset{0}{\overset{2\pi }{\int }}\frac{\cos 6x\cos 7x\cos 8x\cos 9x}{1+e^{{\sin }^{3}4x}}dx\cdots \left(i\right)\{\because \text{Period is}2\pi \}=5\underset{0}{\overset{2\pi }{\int }}\frac{\cos 6x\cos 7x\cos 8x\cos 9x}{1+e^{-{\sin }^{3}4x}}dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$
$\Rightarrow 2I=5\underset{0}{\overset{2\pi }{\int }}\cos 6x\cos 7x\cos 8x\cos 9xdx\Rightarrow I=10\underset{0}{\overset{\pi /2}{\int }}\cos 6x\cos 7x\cos 8x\cos 9xdx\cdots \left(iii\right)\Rightarrow I=10\underset{0}{\overset{\pi /2}{\int }}\cos 6x\sin 7x\cos 8x\sin 9xdx\cdots \left(iv\right)$
Adding $\left(iii\right)$ and $\left(iv\right)$
$\Rightarrow 2I=10\underset{0}{\overset{\pi /2}{\int }}\cos 6x\cos 8x\left(\cos 2x\right)dx\Rightarrow I=5\left(2\underset{0}{\overset{\pi /4}{\int }}\cos 6x\cos 8x\cos 2xdx\right)\cdots \left(v\right)I=10\underset{0}{\overset{\pi /4}{\int }}-\sin 6x\cos 8x\sin 2xdx\cdots \left(vi\right)$
Adding $\left(v\right)$ and $\left(vi\right)$
$\Rightarrow 2I=10\underset{0}{\overset{\pi /4}{\int }}\cos 8x\cos 8xdxI=\frac{5}{2}\underset{0}{\overset{\pi /4}{\int }}(1+\cos 16x)dx=\frac{5}{2}{(x+\frac{\sin 16x}{16})}_0^{\pi /4}=\frac{5\pi }{8}$