JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Definite Integrals

Note that $f(x)=7{\tan }^{6}x({\tan }^{2}x+1)-3{\tan }^{2}x({\tan }^{2}x+1)=7{\tan }^{6}x{\sec }^{2}x-3{\tan }^{2}x{\sec }^{2}x$. Integrating: $\int f(x)dx={\tan }^{7}x-{\tan }^{3}x+C$. Thus ${\int }_0^{\pi /4}f(x)dx=[{\tan }^{7}x-{\tan }^{3}x{]}_0^{\pi /4}=(1-1)-(0-0)=0$ (B is correct). For ${\int }_0^{\pi /4}xf(x)dx$, use integration by parts: $\int xf(x)dx=x({\tan }^{7}x-{\tan }^{3}x)-\int ({\tan }^{7}x-{\tan }^{3}x)dx$. Evaluating at the limits gives $\frac{\pi }{4}(1-1)-0=0$ for the first term, but the second integral needs careful evaluation. Detailed calculation shows ${\int }_0^{\pi /4}xf(x)dx=\frac{1}{12}$ (A is correct).