Evaluate the following. intfrac{sintheta(costheta+4)dtheta}{1+cos^2theta}

Question
Trigonometry
asked 2021-02-09
Evaluate the following.
\(\displaystyle\int{\frac{{{\sin{\theta}}{\left({\cos{\theta}}+{4}\right)}{d}\theta}}{{{1}+{{\cos}^{{2}}\theta}}}}\)

Answers (1)

2021-02-10
Let,
\(\displaystyle{I}=\int{\frac{{{\sin{\theta}}{\left({\cos{\theta}}+{4}\right)}{d}\theta}}{{{1}+{{\cos}^{{2}}\theta}}}}\)
\(\displaystyle{I}=\int{\frac{{{\sin{\theta}}{\left({\cos{\theta}}+{4}\right)}{d}\theta}}{{{1}+{{\cos}^{{2}}\theta}}}}\)
\(\displaystyle=\int{\frac{{{\sin{\theta}}{\cos{\theta}}{d}\theta}}{{{1}+{{\cos}^{{2}}\theta}}}}+\int{\frac{{{4}{\sin{\theta}}{d}\theta}}{{{1}+{{\cos}^{{2}}\theta}}}}\)
\(\displaystyle={I}_{{1}}+{I}_{{2}}\)
Now,
\(\displaystyle{I}_{{1}}=\int{\frac{{{\sin{\theta}}{\cos{\theta}}{d}\theta}}{{{1}+{{\cos}^{{2}}\theta}}}}\)
Put \(\displaystyle{1}+{{\cos}^{{2}}\theta}={t}\Rightarrow{2}{\cos{\theta}}{\left(-{\sin{\theta}}\right)}{d}\theta={\left.{d}{t}\right.}\Rightarrow{\cos{\theta}}{\sin{\theta}}={\frac{{-{1}}}{{{2}}}}{\left.{d}{t}\right.}\)
\(\displaystyle\because{I}_{{1}}=\int{\frac{{-{1}{\left.{d}{t}\right.}}}{{{2}{t}}}}\)
\(\displaystyle={\frac{{-{1}}}{{{2}}}}{\ln{{\left({t}\right)}}}+{C}\)
\(\displaystyle={\frac{{-{\ln{{\left({1}+{{\cos}^{{2}}\theta}\right)}}}}}{{{2}}}}+{C}\)
Now,
\(\displaystyle{I}_{{2}}=\int{\frac{{{4}{\sin{\theta}}{d}\theta}}{{{1}+{{\cos}^{{2}}\theta}}}}\)
Put \(\displaystyle{\cos{\theta}}={t}\Rightarrow-{\sin{\theta}}{d}\theta={\left.{d}{t}\right.}\Rightarrow{\sin{\theta}}{d}\theta=-{\left.{d}{t}\right.}\)
\(\displaystyle\because{I}_{{2}}=\int{\frac{{-{4}}}{{{1}+{t}^{{2}}}}}{\left.{d}{t}\right.}\)
\(\displaystyle=-{4}{{\tan}^{{-{1}}}{\left({t}\right)}}+{d}\)
\(\displaystyle=-{4}{{\tan}^{{-{1}}}{\left({\cos{\theta}}\right)}}+{d}\)
Thus,
\(\displaystyle{I}={I}_{{1}}+{I}_{{2}}\)
\(\displaystyle={\frac{{-{\ln{{\left({1}+{{\cos}^{{2}}\theta}\right)}}}}}{{{2}}}}+{c}-{4}{{\tan}^{{-{1}}}{\left({\cos{\theta}}\right)}}+{d}\)
\(\displaystyle={\frac{{-{\ln{{\left({1}+{{\cos}^{{2}}\theta}\right)}}}}}{{{2}}}}-{4}{{\tan}^{{-{1}}}{\left({\cos{\theta}}\right)}}+{C}\)
0

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