find-dx-cosx-sin-2-x-

Question Number 45795 by maxmathsup by imad last updated on 16/Oct/18

f i n d \int \frac{d x}{c o s x {s i n}^{2} x}

Answered by MJS last updated on 17/Oct/18

\frac{1}{\cos x \sin^{2} x} = (1 + \frac{\cos^{2} x}{\sin^{2} x}) \frac{1}{\cos x} =

= \frac{1}{\cos x} + \frac{\cos x}{\sin^{2} x}

\int \frac{d x}{\cos x} = \int \sec x d x = \ln ∣ \tan x + \sec x ∣ = \ln ∣ \frac{1 + \sin x}{\cos x} ∣

\int \frac{\cos x}{\sin^{2} x} d x =

[t = \sin x \to d x = \frac{d t}{\cos x}]

= \int \frac{d t}{t^{2}} = - \frac{1}{t} = - \frac{1}{\sin x}

\int \frac{d x}{\cos x \sin^{2} x} = \ln ∣ \frac{1 + \sin x}{\cos x} ∣ - \frac{1}{\sin x} + C

Commented by math khazana by abdo last updated on 17/Oct/18

t h a n k y o u s i r M J S .

Commented by MJS last updated on 17/Oct/18

btw

\frac{1}{\cos^{2} x \sin x} = (1 + \frac{\sin^{2} x}{\cos^{2} x}) \frac{1}{\sin x} =

= \frac{1}{\sin x} + \frac{\sin x}{\cos^{2} x}

\int \frac{d x}{\sin x} = \int \csc x d x = - \ln ∣ \cot x + \csc x ∣ = - \ln ∣ \frac{1 + \cos x}{\sin x} ∣

\int \frac{\sin x}{\cos^{2} x} d x =

[t = \cos x \to d x = - \frac{d t}{\sin x}]

= - \int \frac{d t}{t^{2}} = \frac{1}{t} = \frac{1}{\cos x}

Commented by MJS last updated on 17/Oct/18

{you}^{'} re welcome . I just fixed a typo \dots

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