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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)

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

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