Question

Evaluate: ${\int }_0^{\frac{\pi }{4}}({\tan }^{4}x+{\tan }^{2}x)dx$

• A. $1$
• B. $1/2$
• C. $1/3$
• D. $1/4$

Answer 1

Answer: (C)

$1/3$

${\int }_0^{\frac{\pi }{4}}({\tan }^{4}x+{\tan }^{2}x)dx={\int }_0^{\frac{\pi }{4}}{\tan }^{2}x({\tan }^{2}x+1)dx=={\int }_0^{\frac{\pi }{4}}{\tan }^{2}x\left({\sec }^{2}x\right)dx$

We know that:

$\int \left[f\right(x){]}^n{f}^{\prime }\left(x\right)dx=\frac{\left[f\right(x){]}^{n+1}}{n+1}+C={\int }_0^{\frac{\pi }{4}}{\tan }^{2}x\times {\sec }^{2}xdx={\left[\frac{{\tan }^{3}x}{3}\right]}_0^{\frac{\pi }{4}}=\frac{1}{3}\left[{\tan }^3\right(\frac{\pi }{4})-{\tan }^{3}0]=\frac{1}{3}[1-0]=\frac{1}{3}$