Question

${\int }_0^{\frac{\pi }{4}}ta{n}^{2}xdx=$ [Roorkee 1983, Pb. CET 2000]

• A. $1-\frac{\pi }{4}$
• B. $1+\frac{\pi }{4}$
• C. $\frac{\pi }{4}-1$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $1-\frac{\pi }{4}$

${\int }_0^{\frac{\pi }{4}}ta{n}^{2}xdx={\int }_0^{\frac{\pi }{4}}(se{c}^{2}x-1)dx$
$={\int }_0^{\frac{\pi }{4}}se{c}^{2}xdx-{\int }_0^{\frac{\pi }{4}}1dx=[tanx{]}_0^{\frac{\pi }{4}}-[x{]}_0^{\frac{x}{4}}=1-\frac{\pi }{4}$