Mathematics

# If $\displaystyle f(x)=\frac{{x}^{7}-3{x}^{5}+7{x}^{3}-x+1}{{\cos}^{2}x}$ then $\displaystyle \int_{-\pi/4}^{\pi/4}{f(x)}dx$ is equals to

$2$

##### SOLUTION
$f\left( x \right) =\dfrac { { x }^{ 7 }-{ 3x }^{ 5 }+{ 7x }^{ 3 }-x+1 }{ { \cos }^{ 2 }x }$
$\int _{ \dfrac { -\pi }{ 4 } }^{ \dfrac { \pi }{ 4 } }{ f\left( x \right) } dx\quad \quad \left\{ \because \quad f\left( x \right) =-f\left( -x \right) \right\}$
$=\int _{ \dfrac { -\pi }{ 4 } }^{ \dfrac { \pi }{ 4 } }{ \dfrac { dx }{ { \cos }^{ 2 }x } }$
$={ \left( tanx \right) }_{ -\pi /4 }^{ \pi /4 }=2$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Subjective Medium
Evaluate $\displaystyle\int^1_0\dfrac{dx}{(1+x^2)}$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int \dfrac {dx}{(\sqrt {1 + x^{2}} - x)^{n}}(n\neq \pm 1) = \dfrac {1}{2} \left (\dfrac {z^{n + 1}}{n + 1} + \dfrac {z^{n - 1}}{n - 1}\right ) + O$
where
• A. $z = x - \sqrt {1 + x^{2}}$
• B. $z = \sqrt {1 + x^{2}} - x$
• C. $z = x - \sqrt {1 - x^{2}}$
• D. $z = x + \sqrt {1 + x^{2}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int e^{x}\sqrt{1+e^{x}}dx=$
• A. $(1+e^{x})^{\frac{3}{2}}+c$
• B. $\displaystyle \frac{2}{3}(1-e^{x})^{3/{2}}+c$
• C. $(1-e^{x})^{3/{2}}+c$
• D. $\displaystyle \frac{2}{3}(1+e^{x})^{3/{2}}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
$\displaystyle \int { \dfrac { \cos ^{ 2 }{ x } }{ 2+\sin { x } } } dx$

$\int \frac{2x^{2}}{3x^{4}2x} dx$