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  • provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 26

provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 26

Answers (1)

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

Hint:Use substitution method to solve this integral.

Given:   \int \frac{\cos x-\sin x}{1+\sin 2 x} d x

Solution:

        \text { Let } I=\int \frac{\cos x-\sin x}{1+\sin 2 x} d x

                   =\int \frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+\sin 2 x} d x \quad\left[\because 1=\sin ^{2} x+\cos ^{2} x\right]

                   =\int \frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} d x \quad[\because \sin 2 x=2 \sin x \cos x]

                   =\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x \quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right]

        \text { Put } \sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t \text { then }

                I=\int \frac{(\cos x-\sin x)}{t^{2}} \frac{d t}{(\cos x-\sin x)}

                    =\int \frac{1}{t^{2}}=\int t^{-2} d t=\frac{t^{-2+1}}{-2+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

                    =\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c

                    =\frac{-1}{\sin x+\cos x}+c \quad[\because t=\sin x+\cos x]

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