Evaluate the following. intsin^2xcdottan xdx

Evaluate the following. intsin^2xcdottan xdx

Question
Trigonometry
asked 2021-02-24
Evaluate the following.
\(\displaystyle\int{{\sin}^{{2}}{x}}\cdot{\tan{{x}}}{\left.{d}{x}\right.}\)

Answers (1)

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 \(\displaystyle{\frac{{{{\cos}^{{2}}{x}}}}{{{2}}}}-{\ln}{\left|{\cos{{x}}}\right|}+{C}\)
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