Mathematics

# Evaluate $\displaystyle\int_{0}^{\pi/2}\dfrac{\cos x}{1+\sin^{2}x}dx$

##### SOLUTION
$\displaystyle\int_{0}^{\frac{\pi}{2}}{\dfrac{\cos{x}}{1+{\sin}^{2}{x}}dx}$

Let $t=\sin{x}\Rightarrow\,dt=\cos{x}dx$

As $x\rightarrow\,0\Rightarrow\,t\rightarrow\,0$

$x\rightarrow\,\dfrac{\pi}{2}\Rightarrow\,t\rightarrow\,1$

$\displaystyle\int_{0}^{\frac{\pi}{2}}{\dfrac{\cos{x}}{1+{\sin}^{2}{x}}dx}$

$=\displaystyle\int_{0}^{1}{\dfrac{dt}{1+{t}^{2}}}$

$=\left[{\tan}^{-1}{t}\right]_{0}^{1}$

$={\tan}^{-1}{1}-{\tan}^{-1}{0}$

$=\dfrac{\pi}{4}-0$

$=\dfrac{\pi}{4}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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