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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 49

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 49

Answers (1)

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Answer:

                \frac{-1}{27} \log |-\sin x+1|+\frac{1}{9(1-\sin x)}+\frac{1}{6(1-\sin x)^{2}}+\frac{1}{27} \log |2+\sin x|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{\cos x d x}{(1-\sin x)^{3}(2+\sin x)}

Explanation:

Let

I=\int \frac{\cos x d x}{(1-\sin x)^{3}(2+\sin x)}

Let

\begin{aligned} &\sin x=y \\ &\cos x d x=d y \\ &I=\int \frac{d y}{(1-y)^{3}(2+y)} \end{aligned}

Now

\begin{aligned} &\frac{1}{(1-y)^{3}(2+y)}=\frac{A}{1-y}+\frac{B}{(1-y)^{2}}+\frac{C}{(1-y)^{3}}+\frac{D}{2+y} \\ &1=A(1-y)^{2}(2+y)+B(1-y)(2+y)+C(2+y)+D(1-y)^{3} \end{aligned}

\begin{aligned} &\text { Put } y=1 \\ &1=A(0)+B(0)+C(3)+D(0) \\ &1=3 C \end{aligned}

C= \frac{1}{3}                                  (1)

\begin{aligned} &\text { Put } y=-2 \\ &1=A(0)+B(0)+C(0)+D(27) \\ &1=27 D \\ &D=\frac{1}{27} \end{aligned}

 

\begin{aligned} &\text { Similarly } A=\frac{-1}{27}, B=\frac{1}{9} \\ &\frac{1}{(1-y)^{3}(2+y)}=\frac{-1}{27(1-y)}+\frac{1}{9(1-y)^{2}}+\frac{1}{3(1-y)^{3}}+\frac{1}{27(2+y)} \end{aligned}

 

Thus

I=\frac{-1}{27} \int \frac{1}{1-y} d y+\frac{1}{9} \int \frac{1}{(1-y)^{2}} d y+\frac{1}{3} \int \frac{1}{(1-y)^{3}} d y+\frac{1}{27} \int \frac{1}{2+y} d y

\begin{aligned} &I=\frac{-1}{27} \log |1-y|+\frac{1}{9(1-y)}+\frac{1}{6(1-y)^{2}}+\frac{1}{27} \log |2+y|+C \\ &\ldots \ldots \ldots .\left[\int \frac{1}{(1-a)^{2}} d a=\frac{1}{(1-a)}\right] \end{aligned}

I=\frac{-1}{27} \log |1-\sin x|+\frac{1}{9[1-\sin x]}+\frac{1}{6(1-\sin x)}+\frac{1}{27} \log |2+\sin x|+C

Posted by

infoexpert27

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