Mathematics

# Solve:$\int_{0}^{\frac{\pi }{2}}{\dfrac{\sin xdx}{9+{{\cos }^{2}}x}}$

##### SOLUTION

We have,

$\int_{0}^{\frac{\pi }{2}}{\dfrac{\sin xdx}{9+{{\cos }^{2}}x}}$

Let

$\cos x=t$

$\sin xdx=dt$

Therefore,

$\int_{1}^{0}{\dfrac{dt}{9+{{t}^{2}}}}$

$\int_{1}^{0}{\dfrac{dt}{{{t}^{2}}+{{3}^{2}}}}={{\left[ \dfrac{1}{3}{{\tan }^{-1}}\dfrac{t}{3} \right]}_{1}}^{0}$

$=\left[ \dfrac{1}{3}{{\tan }^{-1}}\dfrac{0}{3}-\dfrac{1}{3}{{\tan }^{-1}}\dfrac{1}{3} \right]$

$=\left[ \dfrac{1}{3}{{\tan }^{-1}}\tan 0-\dfrac{1}{3}{{\tan }^{-1}}\dfrac{1}{3} \right]$

$=-\dfrac{1}{3}{{\tan }^{-1}}\dfrac{1}{3}$

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

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