Question

${\int }_0^{\pi /2}\frac{cos2x-cos7x}{1+2cos3x}dx=$

• A. $2$
• B. $0$
• C. $-1$
• D. $1$

Answer 1

Answer: (D)

$1$

$cos2x-cos7x=2sin\frac{9x}{2}sin\frac{5x}{2}$
$1+2cos3x=1+2(1-2si{n}^2\frac{3x}{2})=3-4si{n}^2\frac{3x}{2}$
$\therefore {\int }_0^{\frac{\pi }{2}}\frac{cos2x-cos7x}{1+2cos3x}dx={\int }_0^{\frac{\pi }{2}}\frac{2sin\frac{9x}{2}sin\frac{5x}{2}}{3-4si{n}^2\frac{3x}{2}}dx$
$={\int }_0^{\frac{\pi }{2}}\frac{2sin\frac{9x}{2}sin\frac{5x}{2}.sin\frac{3x}{2}}{3sin\frac{3x}{2}-4si{n}^3\frac{3x}{2}}dx$
$={\int }_0^{\frac{\pi }{2}}\frac{2sin\frac{9x}{2}sin\frac{5x}{2}.sin\frac{3x}{2}}{sin\frac{9x}{2}}dx$
$={\int }_0^{\frac{\pi }{2}}2sin\frac{5x}{2}.sin\frac{3x}{2}dx$
$={\int }_0^{\frac{\pi }{2}}(cosx-cos4x)dx$
$={[sinx-\frac{1}{4}sin4x]}_0^{\frac{\pi }{2}}$
$=1$