∫ ((1+csc 2x)/(1−sin 2x)) dx ?


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Question Number 102303 by bemath last updated on 08/Jul/20

$\int \frac{1 + \csc 2 x}{1 - \sin 2 x} d x ?$
Commented by bemath last updated on 08/Jul/20

$t h a n k y o u b o t h$
	Answered by 1549442205 last updated on 08/Jul/20

	$F = \int \frac{1}{1 - \sin 2 x} dx + \int \frac{\cos 2 x}{{(cosx - sinx)}^{2}} dx$ $\int \frac{dx}{{(cosx - sinx)}^{2}} + \int \frac{cosx + sinx}{cosx - sinx} dx$ $= \int \frac{dx}{\frac{1}{2} \cos^{2} (x + \frac{π}{4})} + \int \frac{\frac{\sqrt{2}}{2} \sin (x + \frac{π}{4}) dx}{\frac{\sqrt{2}}{2} \cos (x + \frac{π}{4})}$ $= 2 \tan (x + \frac{π}{4}) + \int \tan (x + \frac{π}{4}) dx$ $= 2 \tan (x + \frac{π}{4}) + \ln ∣ \cos (x + \frac{π}{4}) ∣ + C$
	Answered by Dwaipayan Shikari last updated on 08/Jul/20
	$∫(1/(1−sin2x))−(1/2)∫((−2cos2x)/(1−sin2x))dx ∫(1/(1−((2t)/(t^2 +1)))).(1/(t^2 +1))dt−(1/2)log(1−sin2x) {put tanx=t ∫(1/((t−1)^2 ))dt−(1/2)log(1−sin2x)=(1/(1−t))−(1/2)log(1−sin2x) =((cosx)/(cosx−sinx))−(1/2)log(1−sin2x)+C$
	$\int \frac{1}{1 - s i n 2 x} - \frac{1}{2} \int \frac{- 2 c o s 2 x}{1 - s i n 2 x} d x$ $\int \frac{1}{1 - \frac{2 t}{t^{2} + 1}} . \frac{1}{t^{2} + 1} d t - \frac{1}{2} l o g (1 - s i n 2 x) {p u t t a n x = t$ $\int \frac{1}{{(t - 1)}^{2}} d t - \frac{1}{2} l o g (1 - s i n 2 x) = \frac{1}{1 - t} - \frac{1}{2} l o g (1 - s i n 2 x)$ $= \frac{c o s x}{c o s x - s i n x} - \frac{1}{2} l o g (1 - s i n 2 x) + C$
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