Mathematics

# The value of integral $\displaystyle \int _{ 0 }^{ \pi }{ \frac { { x }^{ 2 }\sin { x } }{ \left( 2x-\pi \right) \left( 1+\cos ^{ 2 }{ x } \right) } } dx$ is equal to

$\displaystyle\frac { { \pi }^{ 2 } }{ 4 }$

##### SOLUTION
Let $\displaystyle I=\int _{ 0 }^{ \pi }{ \frac { { x }^{ 2 }\sin { x } }{ \left( 2x-\pi \right) \left( 1+\cos ^{ 2 }{ x } \right) } } dx$ $\displaystyle=\int _{ 0 }^{ \pi }{ \frac { { \left( \pi -x \right) }^{ 2 }\sin { { \left( \pi -x \right) } } }{ \left( 2\pi -2x-\pi \right) \left( 1+\cos ^{ 2 }{ { \left( \pi -x \right) } } \right) } } dx$

$\displaystyle=\int _{ 0 }^{ \pi }{ \frac { \left( { \pi }^{ 2 }-2\pi x+{ x }^{ 2 } \right) \sin { { x } } }{ \left( \pi -2x \right) \left( 1+\cos ^{ 2 }{ x } \right) } } dx$ $\displaystyle=\int _{ 0 }^{ \pi }{ \frac { \left( { \pi }^{ 2 }-2\pi x \right) \sin { { x } } }{ \left( \pi -2x \right) \left( 1+\cos ^{ 2 }{ x } \right) } } dx-I$

$\displaystyle\Rightarrow 2I=\pi \int _{ 0 }^{ \pi }{ \frac { \sin { x } }{ 1+\cos ^{ 2 }{ x } } } dx=-\pi \int _{ 1 }^{ -1 }{ \frac { dt }{ 1+{ t }^{ 2 } } }$   [putting $\cos { x } =t\Rightarrow \sin { x } dx=-dt$]

$\displaystyle=\pi { \left[ \tan ^{ -1 }{ t } \right] }_{ -1 }^{ 1 }=\pi \left( \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right) =\frac { { \pi }^{ 2 } }{ 2 }$
$\displaystyle\therefore I=\frac { { \pi }^{ 2 } }{ 4 }$

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

#### Realted Questions

Q1 Subjective Medium
Find
$\int x ^ { 2 } + 2 x d x$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\displaystyle \int_{1}^{2}{x+3}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate:
$\displaystyle\int\limits_{-a}^{a}x^3\sqrt{x^2-a^2}\ dx$.

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Q4 Single Correct Hard
If $\displaystyle A=\int_{0}^{\pi} \dfrac{cos x}{(x+2)^2} \: dx$, then $\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{\sin 2x}{(x+1)} \: dx$ is equal to
• A. $\dfrac{1}{\pi+2}-A$
• B. ${1}+\dfrac{1}{\pi+2}-A$
• C. $A-\dfrac{1}{2}-\dfrac{1}{\pi+2}$
• D. $\dfrac{1}{2}+\dfrac{1}{\pi+2}-A$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$