# Evaluate the following. intsin^2xcdottan xdx

Question
Trigonometry
Evaluate the following.
$$\displaystyle\int{{\sin}^{{2}}{x}}\cdot{\tan{{x}}}{\left.{d}{x}\right.}$$

2021-02-25
The given integral is:
$$\displaystyle\int{{\sin}^{{2}}{x}}\cdot{\tan{{x}}}{\left.{d}{x}\right.}$$
we have to evaluate the given integral.
Let the given integral be I.
Therefore,
$$\displaystyle{I}=\int{{\sin}^{{2}}{x}}\cdot{\tan{{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\left({1}-{{\cos}^{{2}}{x}}\right)}\cdot{\tan{{x}}}{\left.{d}{x}\right.}$$
(because $$\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}$$ therefore $$\displaystyle{{\sin}^{{2}}{x}}={1}-{{\cos}^{{2}}{x}}$$)
$$\displaystyle=\int{\left({1}-{{\cos}^{{2}}{x}}\right)}\cdot{\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}{\left.{d}{x}\right.}$$ (because $$\displaystyle{\tan{{x}}}={\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}$$)
Now let $$\displaystyle{\cos{{x}}}={t}$$
therefore,
$$\displaystyle{d}{\left({\cos{{x}}}\right)}={\left.{d}{t}\right.}$$
$$\displaystyle-{\sin{{x}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}$$
$$\displaystyle{\sin{{x}}}{\left.{d}{x}\right.}=-{\left.{d}{t}\right.}$$
now substitute these values in the integral I.
Therefore,
$$\displaystyle{I}=\int{\left({1}-{t}^{{2}}\right)}\cdot{\frac{{-{\left.{d}{t}\right.}}}{{{t}}}}$$
$$\displaystyle=\int{\left({\frac{{{t}^{{2}}-{1}}}{{{t}}}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={\left({\frac{{{t}^{{2}}}}{{{t}}}}-{\frac{{{1}}}{{{t}}}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle=\int{\left({t}-{\frac{{{1}}}{{{t}}}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle=\int{t}{\left.{d}{t}\right.}-\int{\frac{{{\left.{d}{t}\right.}}}{{{t}}}}$$
$$\displaystyle={\frac{{{t}^{{2}}}}{{{2}}}}-{\ln}{\left|{t}\right|}+{C}$$
where C is the constant of integration
now substitute the value of t that is $$\displaystyle{t}={\cos{{x}}}$$ in the integral I.
therefore,
$$\displaystyle{I}={\frac{{{t}^{{2}}}}{{{2}}}}-{\ln}{\left|{t}\right|}+{C}$$
$$\displaystyle={\frac{{{\left({\cos{{x}}}\right)}^{{2}}}}{{{2}}}}-{\ln}{\left|{\cos{{x}}}\right|}+{C}$$
$$\displaystyle{\frac{{{{\cos}^{{2}}{x}}}}{{{2}}}}-{\ln}{\left|{\cos{{x}}}\right|}+{C}$$
therefore,
$$\displaystyle{I}=\int{{\sin}^{{2}}{x}}\cdot{\tan{{x}}}{\left.{d}{x}\right.}={\frac{{{{\cos}^{{2}}{x}}}}{{{2}}}}-{\ln}{\left|{\cos{{x}}}\right|}+{C}$$
therefore the value of the given integral $$\displaystyle\int{{\sin}^{{2}}{x}}\cdot{\tan{{x}}}{\left.{d}{x}\right.}$$ is $$\displaystyle{\frac{{{{\cos}^{{2}}{x}}}}{{{2}}}}-{\ln}{\left|{\cos{{x}}}\right|}+{C}$$

### Relevant Questions

Evaluate the following
$$\displaystyle\int{{\csc}^{{6}}{u}}{d}{u}$$
Evaluate the following.
$$\displaystyle\int{\cos{\beta}}{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}{d}\beta$$
Evaluate the following.
$$\displaystyle\int{\frac{{{\left({10}{y}+{11}\right)}{\left.{d}{y}\right.}}}{{{4}{y}^{{2}}-{4}{y}+{5}}}}$$
Evaluate the following.
$$\displaystyle\int{\frac{{{{\cos}^{{5}}{\left({3}{z}\right)}}{\left.{d}{z}\right.}}}{{{{\sin}^{{2}}{\left({3}{z}\right)}}}}}$$
Evaluate the following.
$$\displaystyle\int{\frac{{{\sin{\theta}}{\left({\cos{\theta}}+{4}\right)}{d}\theta}}{{{1}+{{\cos}^{{2}}\theta}}}}$$
Explain how we can evaluate the expression $$\displaystyle{\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}$$ and what the evaluation is in terms of v and a.
$$\displaystyle{{\tan{{50}}}^{\circ}}$$
$$\displaystyle{\sec{{\left({210}\right)}}}\times{\cot{{\left({300}\right)}}}+{\sin{{\left({225}\right)}}}$$
$$\displaystyle{\tan{{\left({60}\right)}}}\times{3}{\sin{{\left({90}\right)}}}-{\sin{{\left({315}\right)}}}$$
$$\displaystyle{\tan{{\left({330}\right)}}}\times{\cos{{\left({240}\right)}}}-{2}{\cos{{\left({270}\right)}}}$$