Question

Assertion :${\int }_0^{\pi }x\tan x{\cos }^{3}xdx=\frac{\pi }{2}{\int }_0^{\pi }\tan x{\cos }^{3}xdx.$ Reason: ${\int }_a^{b}xf\left(x\right)dx=\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx.$

• 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. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Assertion
$l={\int }_0^{n}x\tan x{\cos }^{3}xdx$ ....(i)
$l={\int }_0^n(\pi -x)\tan x{\cos }^{3}xdx$...(ii)
(i)+(ii)
$2l=\pi {\int }_0^{\pi }\tan x\cdot {\cos }^3$
$l=\frac{\pi }{2}{\int }_0^{\pi }\tan x{\cos }^{3}xdx$
Reason
$l={\int }_a^{b}xf\left(x\right)dx$..(i)
$l={\int }_a^b(a+b-x)f(a+b-x)dx$...(ii)
(i)+(ii)
$2l=(a+b){\int }_a^{b}f\left(x\right)dx$ if f(a+b-x)=f(x)
$l=\frac{a+b}{2}{\int }_b^{a}f\left(x\right)dx$
Hence reason is false but if $f(a+b-x)\ne f\left(x\right)$ then
$l\ne \frac{a+b}{2}{\int }_a^{b}f\left(x\right)$