Question

Assertion :The value of ${\int }_0^{\pi /2}{\sin }^{6}xdx=\frac{5\pi }{16}$ Reason: If n is even, then ${\int }_0^{\pi /2}{\sin }^{n}xdx$ equals $\frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}...\frac{1}{2}\times \frac{\pi }{2}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct.
  

Answer 1

The correct option is D Assertion is incorrect but Reason is correct.

$I={\int }_0^{\frac{\pi }{2}}si{n}^6\left(x\right).dx$
$={\int }_0^{\pi }2si{n}^6(\frac{\pi }{2}-x).dx$
$={\int }_0^{\frac{\pi }{2}}co{s}^6\left(x\right).dx$
Hence
$2I={\int }_0^{\frac{\pi }{2}}si{n}^6\left(x\right)+co{s}^6\left(x\right).dx$
$si{n}^6\left(x\right)+co{s}^6\left(x\right)=\left[\right(si{n}^2\left(x\right)+co{s}^2\left(x\right)\left)\right(si{n}^4\left(x\right)+co{s}^4\left(x\right)-si{n}^2\left(x\right).co{s}^2\left(x\right)\left)\right]$
$=si{n}^4\left(x\right)+co{s}^4\left(x\right)-si{n}^2\left(x\right).co{s}^2\left(x\right))$
$=\left(si{n}^2\right(x)+co{s}^2{)}^2-2si{n}^{2}x.co{s}^{2}x-si{n}^2(x\left)co{s}^2\right(x)$
$=1-3si{n}^{2}x.co{s}^{2}x$
$=1-\frac{3si{n}^{2}2\left(x\right)}{4}$
$=1-\frac{3(1-cos4x)}{8}$
$=\frac{5}{8}+\frac{3cos4x}{8}$
Hence
$2I={\int }_0^{\frac{\pi }{2}}\frac{5}{8}+\frac{3cos4x}{8}.dx$
$2I=\frac{5\pi }{16}$
$I=\frac{5\pi }{32}$.
Hence Assertion is false.