Mathematics

# Solve $\displaystyle\int\dfrac{{\cos x - \sin x}}{{1 + \sin 2x}}dx$

##### SOLUTION

Solve the given trigonometric function.

$=\displaystyle\int\dfrac{cosx-sinx}{1+sin2x} dx$

$=\displaystyle\int\dfrac{cosx-sinx}{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x} dx$

$=\displaystyle\int\dfrac{cosx-sinx}{{{\left( cosx+sinx \right)}^{2}}} dx$

Let $cosx+sinx=t$

Then,

$\dfrac{d}{dx}\left( cosx+sinx \right)=\dfrac{d}{dx}t$

$\Rightarrow -\sin x+\cos x=\dfrac{dt}{dx}$

$\Rightarrow \left( \cos x-\sin x \right)dx=dt$

Now,

$\int{\dfrac{dt}{{{t}^{2}}}}$

$=\int{{{t}^{-2}}dt}$

$=\dfrac{{{t}^{-2+1}}}{-2+1}$

$=-\dfrac{1}{t}=-\dfrac{1}{\cos x+\sin x}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

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Let $I_n=\int \tan^n x\,dx, (n > 1)$. If $I_4 +I_6=a \tan^5x+bx^5+C$, where $C$ is a constant of integration, then the ordered pair $(a, b)$ is equal to
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