Mathematics

# Solve: $\int\dfrac{1}{(\sin x-2 \cos x)(2\sin x+\cos x)}dx$

##### SOLUTION
Let, $I=\int { \dfrac { 1 }{ (\sin { x } -2\cos { x) } (2\sin { x } +\cos { x } ) } } dx\\ =\int { \dfrac { 1 }{ 2{ \sin }^ { 2 }x-3\sin { x\cos { x } } -2{ \cos } ^{ 2 }x } } dx\\ =\int { \dfrac { 1 }{ -2({ \cos }^ { 2 }x-{ \sin }^ { 2 }x)-3\sin { x\cos { x } } } } dx\\ =-\int { \dfrac { 1 }{ 2\cos { 2x } +\dfrac { 3 }{ 2 } \sin { 2x } } } dx$
Now, let $\tan { x } =t\\ \therefore dx=\dfrac { 2 }{ 1+{ t }^{ 2 } } dt,\cos { 2x } =\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } ,\sin { 2x } =\dfrac { 2t }{ 1+{ t }^{ 2 } }$
We can write,
$I=-\int { \dfrac { 1 }{ 2(\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } )+\dfrac { 3 }{ 2 } (\dfrac { 2t }{ 1+{ t }^{ 2 } } ) } } \cdot (\dfrac { 2 }{ 1+{ t }^{ 2 } } dt)\\ =-\int { \dfrac { 1 }{ \dfrac { 2-2{ t }^{ 2 }+3t }{ 1+{ t }^{ 2 } } } } \cdot \dfrac { 2 }{ 1+{ t }^{ 2 } } dt\\ =\int { \dfrac { 2 }{ 2{ t }^{ 2 }-3t-2 } } dt\\ =\int { \dfrac { 1 }{ { t }^{ 2 }-\dfrac { 3 }{ 2 } t-1 } } dt\\ =\int { \dfrac { 1 }{ { t }^{ 2 }-\dfrac { 3 }{ 2 } t-1+\dfrac { 9 }{ 16 } -\dfrac { 9 }{ 16 } } } dt\\ =\int { \dfrac { 1 }{ ({ t }^{ 2 }-\dfrac { 3 }{ 2 } t+\dfrac { 9 }{ 16 } )-(\dfrac { 9 }{ 16 } +\dfrac { 16 }{ 16 } ) } } dt\quad \quad \left[ putting,1=\dfrac { 16 }{ 16 } \right] \\ =\int { \dfrac { 1 }{ ({ t-\dfrac { 3 }{ 2 } })^{ 2 }-{ (\dfrac { 5 }{ 4 } ) }^{ 2 } } } dt$
We know that,
$\int { \dfrac { 1 }{ ({ { x }^{ 2 }-{ a }^{ 2 }) } } } dx=\dfrac { 1 }{ 2a } \log { \left[ \dfrac { x-a }{ x+a } \right] } +C$
$\therefore I=\dfrac { 1 }{ 2(\dfrac { 5 }{ 4 } ) } \log { \left[ \dfrac { (t-\dfrac { 3 }{ 2 } )-\dfrac { 5 }{ 4 } }{ (t-\dfrac { 3 }{ 2 } )+\dfrac { 5 }{ 4 } } \right] } +C\\ =\dfrac { 2 }{ 5 } \log { \left[ \dfrac { 4t-11 }{ 4t-1 } \right] } +C$
Putting $t=\tan { x } ,\\ \Rightarrow I=\dfrac { 2 }{ 5 } \log { \left[ \dfrac { 4\tan { x } -11 }{ 4\tan { x } -1 } \right] } +C\\ \therefore \int { \dfrac { 1 }{ (\sin { x } -2\cos { x) } (2\sin { x } +\cos { x } ) } } dx=\dfrac { 2 }{ 5 } \log { \left[ \dfrac { 4\tan { x } -11 }{ 4\tan { x } -1 } \right] } +C.$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
Evaluate: $\displaystyle \int x^{x}(1 + \log x)dx$
• A. $x^{-x} + x$
• B. $x \log x + x$
• C. $\log x + x$
• D. $x^{x} + C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int x^{2}(\log x)^{3}dx=$
• A. $\displaystyle x^{3}\left[ \frac { (\log x)^{ 3 } }{ 3 } +\frac { (\log x)^{ 2 } }{ 3 } +\frac { 2(\log x) }{ 9 } +\frac { 2 }{ 27 } \right] +c$
• B. $\displaystyle x^{3}\left[ \frac { (\log x)^{ 3 } }{ 3 } -\frac { (\log x)^{ 2 } }{ 27 } \right] +c$
• C. $\displaystyle x^{3}\left[ \frac { (\log x)^{ 3 } }{ 3 } +\frac { (\log x)^{ 2 } }{ 3 } +\frac { 2(\log x) }{ 9 } -\frac { 2 }{ 27 } \right] +c$
• D. $\displaystyle x^{3}\left[ \frac { (logx)^{ 3 } }{ 3 } -\frac { (logx)^{ 2 } }{ 3 } +\frac { 2(logx) }{ 9 } -\frac { 2 }{ 27 } \right] +c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate:

$\displaystyle \int {\dfrac{{{x^3} - 4{x^2} + 6x + 5}}{{{x^2} - 2x + 3}}} dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate:
$\displaystyle\int_{-2}^{3}{\left|1-{x}^{2}\right|dx}$
• A. $\dfrac{2}{3}$
• B. $\dfrac{8}{3}$
• C. $\dfrac{5}{3}$
• D. $\dfrac{28}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$