Mathematics

Evaluate: $$\displaystyle\int {\dfrac{1}{{\sin x{{\cos }^2}x}}dx} $$


SOLUTION
$$\displaystyle \int \dfrac{1}{\sin x \cos^2 x} dx$$
$$= \displaystyle\int \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos^2 x} dx$$     [using $$\sin^2 x + \cos^2 x = 1$$]
$$= \displaystyle\int \dfrac{\sin x}{\cos^2 x} dx + \int \dfrac{1}{\sin x} dx$$
for the first integral, put $$\cos x = t$$
$$\Rightarrow - \sin x \, dx = dt$$
$$\Rightarrow dx = \dfrac{-1}{\sin x} dt$$
$$= \displaystyle\int \dfrac{-dt}{t^2} + \int cosec \, x \, dx$$
$$= \dfrac{-(t)^{-2 + 1}}{-2 + 1} + (-ln |cosec \, x + \cot x|)$$
$$= \dfrac{1}{t} - (ln |cosec x + cot x|)$$
$$= \dfrac{1}{\cos x} - ln |cosec x + cot x|$$
$$= \sec x - ln |cosec x + \cot x|$$
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Subjective Medium Published on 17th 09, 2020
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