Choose the correct answer-a frac-1sin- x-cos- x-Cblog -sin- x-cos- x-Cclog -sin- x-cos- x-Cd -frac1sin- x-cos- x- 2-C

Let I=∫cos 2x/(sin x+cos x)^2dx=∫cos^2x sin^2x/(sin x+cos x)^2dx [∵ cos 2x=cos^2x sin^2x] =∫(cos x sin x)(cos x+sin x)/(sin x+cos x)(sin x+cos x)dx ...

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Standard XII

Mathematics

Integration of Piecewise Continuous Functions

Choose the co...

Question

Choose the correct answer.

$\int \frac{c o s 2 x}{(s i n x + c o s x)^{2}} d x i s e q u a l t o (a) \frac{- 1}{s i n x + c o s x} + C (b) l o g | s i n x + c o s x | + C (c) l o g | s i n x - c o s x | + C (d) \frac{1}{(s i n x + c o s x)^{2}} + C$

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Solution

$L e t I = \int \frac{c o s 2 x}{(s i n x + c o s x)^{2}} d x = \int \frac{c o s^{2} x - s i n^{2} x}{(s i n x + c o s x)^{2}} d x [∵ c o s 2 x = c o s^{2} x - s i n^{2} x] = \int \frac{(c o s x - s i n x) (c o s x + s i n x)}{(s i n x + c o s x) (s i n x + c o s x)} d x = \int \frac{c o s x - s i n x}{s i n x + c o s x} d x P u t s i n x + c o s x = t \Rightarrow (c o s x - s i n x d x = d t) ∴ I = \int \frac{c o s x - s i n x}{t} \frac{d t}{c o s x - s i n x} = \int \frac{1}{t} d t = l o g | t | + C = l o g | s i n x + c o s x | + C .$

Hence, the option (b) is correct.

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Choose the correct answer. ∫cos 2x(sin x+cos x)2dx is equal to(a)−1sin x+cos x+C(b)log|sin x+cos x|+C(c)log|sin x−cos x|+C(d)1(sin x+cos x)2+C

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