Find the integrals of the functions-cos 2 x-cos x-sin x2 d x

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

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Standard XII

Mathematics

Basic Trigonometric Identities

Find the inte...

Question

Find the integrals of the functions.
$\int \frac{c o s 2 x}{(c o s x + s i n x)^{2}} d x .$

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Solution

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

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Find the integrals of the functions. ∫cos2x(cosx+sinx)2dx.

Find the integrals of the functions.
$\int \frac{c o s 2 x}{(c o s x + s i n x)^{2}} d x .$