Question

integrate (cos x- sin x)/(1+sin 2x)​

Answer 1

Answer:

$\sf I= -\dfrac{1}{sin\:x+cos\:x} +C$

Step-by-step explanation:

Given:

$\sf \dfrac{cos\:x-sin\:x}{1+sin\:2x}$

To Find:

$\displaystyle \sf \int\limits {\dfrac{cos\:x-sin\:x}{1+sin\:2x} } \, dx$

Solution:

$\displaystyle \sf \int\limits {\dfrac{cos\:x-sin\:x}{1+sin\:2x} } \, dx$

We know that,

sin 2x = 2 sinx cosx

1 = sin²x + cos² x

Substituting this we get,

$\displaystyle \sf \int\limits {\dfrac{cos\:x-sin\:x}{sin^2\:x+cos^2\:x+2\:sin\:cos\:x} } \, dx$

The denominator is the expansion of (sin x + cos x)².

Hence,

$\displaystyle \sf \int\limits {\dfrac{cos\:x-sin\:x}{(sin\:x+cos\:x)^2} } \, dx$

Now let us assume that sin x + cos x = t

Differentiating on both sides we get,

cos x - sin x dx = dt

Substituting in the above integral,

$\displaystyle \sf \int\limits {\dfrac{1}{t^2} } \, dt$

$\sf \implies \dfrac{t^{-2+1}}{-2+1} +C$

$\sf \implies -\dfrac{1}{t} +C$

Give back the value of t,

$\sf \implies -\dfrac{1}{sin\:x+cos\:x} +C$

This is the required integral.