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,

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

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

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

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

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


Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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