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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefiite Integrals Excercise Fill in the Blanks Question 13

Answers (1)

Answer:

\frac{-1}{\sin x+\cos x}+c

Hint:

Simplify it and then integrate it.

Solution:

\begin{aligned} &I=\int \frac{\cos x-\sin x}{1+\sin 2 x} d x \\ &I=\int \frac{\cos x-\sin x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x} d x \quad\quad\quad\quad\quad\quad\quad\quad\left[\cos ^{2} x+\sin ^{2} x=1, \quad \sin 2 x=2 \sin x \cos x\right] \\ &I=\int \frac{\cos x-\sin x}{(\cos x+\sin x)^{2}} d x \end{aligned}

 

Let  \cos x+\sin x=y           …(i)

\begin{aligned} &(-\sin x+\cos x) d x=d y \\ &\Rightarrow I=\int \frac{d y}{y^{2}} \\ &I=\frac{-1}{y}+c \\ &I=\frac{-1}{\sin x+\cos x}+c \end{aligned}

 

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