Mathematics

# Solve:$\displaystyle \int \dfrac{sin x. \cos x}{a\cos^2x+b\sin^2 x}.dx$

##### SOLUTION
Given the integral,
$\int { \dfrac { \cos { x } \sin { x } }{ a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x } } } dx$
Let us assume,
$u=b\sin ^{ 2 }{ x } +a\cos ^{ 2 }{ x } \\ \Rightarrow \dfrac { du }{ dx } =2b\cos { x } \sin { x } -2a\cos { x } \sin { x } \\ \Rightarrow \dfrac { du }{ dx } =\cos { x } \sin { x } (2b-2a)\\ \Rightarrow \cos { x } \sin { x } dx=\dfrac { 1 }{ 2b-2a } du$
Substituting these values in the given integral we get,
$\int { \dfrac { \cos { x } \sin { x } }{ a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x } } } dx\\ =\int { \dfrac { 1 }{ (2b-2a)u } du } \\ =\dfrac { 1 }{ 2b-2a } \int { \dfrac { 1 }{ u } du } \\ =\dfrac { 1 }{ 2b-2a } \ln { (u) } \\ =\dfrac { \ln { (u) } }{ 2b-2a } \\ =\dfrac { \ln { (a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x } ) } }{ 2b-2a } \\ \therefore \int { \dfrac { \cos { x } \sin { x } }{ a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x } } } dx=\dfrac { \ln { \left( \left| a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x } \right| \right) } }{ 2b-2a } +C.$

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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
Partial fraction from of $\displaystyle \frac{3x+7}{x^2-3x+2}$ is
• A. $\displaystyle \frac{13}{x-2}+\frac{10}{x-1}$
• B. $\displaystyle -\frac{13}{x-2}+\frac{10}{x-1}$
• C. $\displaystyle \frac{11}{x-2}-\frac{10}{x-1}$
• D. $\displaystyle \frac{13}{x-2}-\frac{10}{x-1}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate:
$\displaystyle \int { \dfrac { x-{ x }^{ 2 } }{ { x }^{ 2 }-2x-3 } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Evaluate $\displaystyle \int _ { 0 } ^ { \infty } \frac { x ^ { 2 } + 1 } { x ^ { 4 } + 7 x ^ { 2 } + 1 } d x$
• A. $\pi$
• B. $\dfrac { \pi } { 2 }$
• C. $\dfrac { \pi } { 6 }$
• D. $\dfrac { \pi } { 3 }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Prove that $\int\limits_0^{\tfrac {32\pi} 3} {\sqrt {1 + \cos 2x} \,dx = \sqrt {\frac{3}{2}} }$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$