Question

The value of the definite integral ${\int }_0^{\pi /2}({\cos }^{10}x\cdot \sin 12x)dx$, is equal to.

• A. $\frac{1}{10}$
• B. $\frac{1}{11}$
• C. $\frac{1}{12}$
• D. $\frac{1}{22}$

Answer 1

Answer: (B)

$\frac{1}{11}$

Let $I={\int }_0^{\pi /2}({\cos }^{10}x.\sin 12x)dx$

${\cos }^{10}x.\sin 12x={\cos }^{10}x.\sin (11x+x)={\cos }^{10}x.(\sin 11x.\cos x+\cos 11x.\sin x)$

$=-\frac{1}{11}(-11\sin 11x.{\cos }^{11}x-11\cos 11x.{\cos }^{10}x.\sin x)$

$=-\frac{1}{11}\times$ $\frac{d}{dx}(\cos 11x.{\cos }^{11}x)$ ... (1)

Therefore, $I=(-\frac{1}{11}){\int }_0^{\pi /2}\frac{d}{dx}(\cos 11x.{\cos }^{11}x)dx$

$I=-\frac{1}{11}\times \cos 11x.{\cos }^{11}x{]}_0^{\pi /2}$

$I=-\frac{1}{11}\times (0-1)$

$I=\frac{1}{11}$

Correct answer is option B.