Mathematics

# Evaluate : $\int \frac{1}{sinx-sin2x}dx$

##### SOLUTION
$I=\int \dfrac{1}{sinx-sin2x}dx$

$I=\int \dfrac{1}{sinx-2sinxcosx}dx$

$I=\int \dfrac{1}{sinx(1-2cosx)}dx$

$I=\int \dfrac{sinx}{sin^{2}x(1-2cosx)}dx$

$I=\int \dfrac{sinx}{(1-cos^{2}x)(1-2cosx)}dx$

Let $cosx=t$
$-sinxdx=dt$
$sinxdx=-dt$

$I=\int \dfrac{-dt}{(1-t^{2})(1-2t)}$

$I=\int \dfrac{-dt}{(1-t)(1+t)(1-2t)}$

$I=\int \dfrac{1}{(t-1)(t+1)(1-2t)}dt$

Now,
$\dfrac{1}{(t-1)(t+1)(1-2t)}=\dfrac{A}{t+1} +\dfrac{B}{t-1} +\dfrac{C}{1-2t}$

Solving this, we get,
$A=-\dfrac{1}{6}, B=-\dfrac{1}{2}, C=-\dfrac{4}{3}$
Therefore,
$I=\int \dfrac{A}{t+1}dt +\int \dfrac{B}{t-1}dt + \int \dfrac{C}{1-2t}dt$

$I=Alog(t+1)+Blog(t-1)-\dfrac{C}{2}log(1-2t)+C$

$I=\dfrac{-1}{6}log(cosx+1)-\dfrac{1}{2}log(cosx-1)+\dfrac{4}{6}log(1-2cosx)+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle \int\frac{x\cos x\log x-\sin x}{x(\log x)^{2}}dx=$
• A. $\displaystyle \frac { \cos { x } }{ \log { x } } +C$
• B. $\displaystyle \frac { \log { x } }{ \sin { x } } +C$
• C. $\displaystyle \frac { \log { x } }{ \cos { x } } +C$
• D. $\displaystyle \frac { \sin { x } }{ \log { x } } +C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium

Area bounded by $\mathrm{y}=\{\mathrm{x}\},\{.\}$ is fractional part of function and $\mathrm{x}=\pm 1$ is in sq. units
• A. 2
• B. 3
• C. 4
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Q3 Subjective Medium
Write a value of
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1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Solve $\displaystyle\int\limits_1^3 {\frac{{\log x}}{{{{\left( {x + 1} \right)}^2}}}dx}$
• A. $I=\dfrac{3\log 3}{4}+\log \left( 2 \right)$
• B. $I=\dfrac{\log 3}{4}-\log \left( 2 \right)$
• C. None of these
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