Mathematics

# $\int_\limits{0}^{\pi/2}\frac{sinx - cosx}{1+ sinx cosx}dx$

##### SOLUTION

$\int_{0}^{(\frac{\pi}{2})}(\frac{sin(\frac{\pi}{2}-x)-cosx(\frac{\pi}{2}-x)}{1+sin(\frac{\pi}{2}-x) cosx(\frac{\pi}{2}-x)})dx\\=\int_{0}^{(\frac{\pi}{2})}(\frac{cosx-sinx}{1+cosx sinx})dx \\=-\int_{0}^{(\frac{\pi}{2})}(\frac{cosx-sinx}{1+cosx sinx})dx\\ \therefore I =-I\\ \therefore 2I=0 \>then \>I =0$

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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 Hard
$\int e^{x^4}\, (x\, +\, x^3\, +\, 2x^5)e^{x^2}dx$ is equal to
• A. $\displaystyle \frac{1}{2} xe^{x^2}\, \cdot\, e^{x^4}\, +\, c$
• B. $\displaystyle \frac{1}{2} x^2\, e^{x^4}\, +\, c$
• C. $\displaystyle \frac{1}{2} e^{x^2}\, \cdot\, e^{x^4}\, +\, c$
• D. $\displaystyle \frac{1}{2} x^2 e^{x^2}\, \cdot\, e^{x^4}\, +\, c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
lf $\displaystyle \int\frac{2x^{2}+a^{2}}{x^{2}(x^{2}+a^{2})}dx=\frac{k}{x}+\frac{1}{a}tan^{-1}\frac{x}{a}+c$, then k=
• A. $0$
• B. $1$
• C. $\dfrac{1}{a}$
• D. $-1$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Integrate $\displaystyle \int { \cfrac { dx }{ { x }^{ 4 }+1 } }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Integrate with respect to $x$:
$2x^{2}e^{x^{2}}$

Solve $\displaystyle \int\sqrt{\dfrac{a-x}{a+x}}dx$