Mathematics

# $\int _ { 0 } ^ { 4 } \frac { \sin x ^ { 2 } d x } { \sin ( x - 4 ) ^ { 2 } + \sin x ^ { 2 } }$ is equal to

2

##### SOLUTION
$I=\int _{ 0 }^{ 4 }{ \frac { \sin { { x }^{ 2 } } }{ \sin { { \left( x-4 \right) }^{ 2 } } +\sin { { x }^{ 2 } } } dx } \\ I=\int _{ 0 }^{ 4 }{ \frac { \sin { { \left( 4-x \right) }^{ 2 } } }{ \sin { { x }^{ 2 } } +\sin { { \left( 4-x \right) }^{ 2 } } } dx } \quad \quad \quad \left( \int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( a+b-x \right) dx } \right) \\ 2I=\int _{ 0 }^{ 4 }{ \left( \frac { \sin { { x }^{ 2 } } }{ \sin { { \left( x-4 \right) }^{ 2 } } +\sin { { x }^{ 2 } } } +\frac { \sin { { \left( 4-x \right) }^{ 2 } } }{ \sin { { x }^{ 2 } } +\sin { { \left( 4-x \right) }^{ 2 } } } \right) } dx\\ \quad =\int _{ 0 }^{ 4 }{ dx } =4\\ \Rightarrow I=2$
so option A is correct

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

#### Realted Questions

Q1 Single Correct Hard
If $\displaystyle \int _{ 0 }^{ \infty }{ { e }^{ -ax }dx } =\frac { 1 }{ a }$, then $\displaystyle \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx }$ is
• A. $\displaystyle \frac { { \left( -1 \right) }^{ n }n! }{ { a }^{ n+1 } }$
• B. $\displaystyle \frac { { \left( -1 \right) }^{ n }\left( n-1 \right) ! }{ { a }^{ n } }$
• C. None of these
• D. $\displaystyle \frac { n! }{ { a }^{ n+1 } }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate:
$\int { \cfrac { 8 }{ (x+2)({ x }^{ 2 }+4) } dx }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Integrate
$\int {{x \over {{x^2} + x + 1}}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Multiple Correct Hard
$\int_{0}^{2\pi}sin^{4}$ x dx is equal to
• A. $8\int_{0}^{\frac{\pi}{4}}sin^{4}$ x dx
• B. $3\int_{0}^{\frac{2\pi}{3}}sin^{4}$ x dx
• C. $2\int_{0}^{\pi}sin^{4}$ x dc
• D. $4\int_{0}^{\frac{\pi}{2}}cos^{4}$ x dx

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$