Computer Science, asked by abhijeetvshkrma, 4 months ago

Solve this integral...​

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Answers

Answered by BrainlyIAS
11

Question :

\displaystyle \sf \red{\int \dfrac{d \theta}{(sin\ \theta - 2 cos\ \theta)(2\ sin\ \theta + cos\ \theta )}}

Solution :

\\ \to \displaystyle \sf \int \dfrac{d \theta}{ \bigg( \frac{cos\ \theta}{cos\ \theta}  \bigg) (sin\ \theta - 2 cos\ \theta) \bigg( \frac{cos\ \theta}{cos\ \theta} \bigg) (2\ sin\ \theta + cos\ \theta )} \\

\\ \to \displaystyle \sf \int \dfrac{d \theta }{(cos^2 \theta )(tan\ \theta - 2)(2\ tan\ \theta + 1)} \\

\\ \bullet\ \; \pink{ \sf sec^2 \theta = \dfrac{1}{ cos^2 \theta }} \\

\\ \to \displaystyle \sf \int \dfrac{sec^2 \theta\ d \theta }{(tan\ \theta - 2)(2\ tan\ \theta + 1)} \\

Let's use substitution ,

u = tan θ

du = sec²θ dθ

\\ \to \displaystyle \sf \int \dfrac{du}{(u-2)(2u+1)} \\

Use partial faction to evaluate , ( Find attachment )

\to \displaystyle \sf \dfrac{1}{5} \int \dfrac{du}{u-2} - \dfrac{2}{5} \int \dfrac{du}{2u+1}

\to \displaystyle \sf \dfrac{1}{5} . \ln (u-2) - \dfrac{2}{5} . \dfrac{ \ln (2u+1)}{2}+c

\to \sf \dfrac{1}{5} \ln (u - 2) - \dfrac{1}{5} \ln (2\ u + 1)+c

\to \sf \dfrac{1}{5} \bigg( \ln (u -2) - \ln (2u + 1) \bigg)+c

\bullet\ \; \sf \pink{ \ln A - \ln B = \dfrac{ \ln A}{ \ln B}}

\to \sf \dfrac{1}{5}\ ln\bigg[ \dfrac{u-2}{ 2u+1} \bigg]+c

  • u = tan θ

\to \sf \green{\dfrac{1}{5}\ \ln\bigg[ \dfrac{  tan\ \theta -2}{ 2\ tan\ \theta+1} \bigg]+c }

★ ═════════════════════ ★

\displaystyle \sf \int \dfrac{d \theta}{(sin\ \theta - 2 cos\ \theta)(2\ sin\ \theta + cos\ \theta )} = \dfrac{1}{5}\ \ln\bigg[ \dfrac{ tan\ \theta -2}{ 2\ tan\ \theta+1} \bigg]+c

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Answered by ramesh015
1

Answer:

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