Question

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

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

Short answer

The answer is $-1+\sqrt{2}$.

Step-by-step solution

Step 1. Given information

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

Step 2. Explanation.

To solve the integral use substitution method.

Substitute $u=\cos x$so

$$\begin{gathered} du=-\sin xdx \\ -du=\sin xdx \end{gathered}$$

So the integral becomes,

${\int }_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}-\frac{1}{u^2}du$

Solve the integral as follows,

$$\begin{gathered} {\int }_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}-\frac{1}{u^2}du=-{\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}^1-\frac{1}{u^2}du \\ ={\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}^1\frac{1}{u^2}du \\ ={\left[-\frac{1}{u}\right]}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}^1 \\ =-1+\sqrt{2} \end{gathered}$$