Question

Solve each of the definite integrals in Exercises 67–76.

${\int }_0^{\mathrm{\pi }/4}\frac{{\sin }^{2}x}{{\cos }^{4}x}dx$.

Short answer

The answer is$\frac{1}{3}$.

Step-by-step solution

Step 1. Given Information.

The integral is${\int }_0^{\mathrm{\pi }/4}\frac{{\sin }^{2}x}{{\cos }^{4}x}dx$.

Step 2. Explanation.

Simplify the integral using trigonometric identities, $\tan x=\frac{\sin x}{\cos x}$and $secx=\frac{1}{\cos x}$.

${\int }_0^{\mathrm{\pi }/4}{\tan }^{2}xse{c}^{2}xdx$

Substitute$u=\tan x$so, $du=se{c}^{2}xdx$

so, limits changes 0 to 1.

role="math" localid="1649243192222" ${\int }_0^1{u}^{2}du$.

Step 3. Calculation.

Further simplify.

$$\begin{gathered} {\int }_0^1{u}^{2}du={\left[\frac{u^3}{3}\right]}_0^1 \\ =\frac{1}{3} \end{gathered}$$