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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 59

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 59

Answers (1)

Answer:

                \frac{1}{18} \log (1+\sin x)-\frac{1}{2} \log (1-\sin x)+\frac{4}{9} \log \left(\frac{5}{4}-\sin x\right)+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{1}{\cos x(5-4 \sin x)} d x

Explanation:

Let

I=\int \frac{1}{\cos x(5-4 \sin x)} d x

\begin{aligned} &\text { Put }\\ &\sin x=t\\ &\cos x d x=d t\\ &d x=\frac{1}{\cos x} d t \end{aligned}

 

\begin{aligned} &I=\int \frac{1}{\cos ^{2} x(5-4 t)}=\int \frac{1}{\left(1-\sin ^{2} x\right)(5-4 t)} d t \\ &I=\int \frac{1}{\left(1-t^{2}\right)(5-4 t)} d t \\ &I=\frac{1}{4} \int \frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)} d t \end{aligned}                        (1)

 

\frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{\frac{5}{4}-t}

\begin{aligned} &\frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)}=\frac{\left[A(1-t)\left(\frac{5}{4}-t\right)+B(1+t)\left(\frac{5}{4}-t\right)+C\left(1-t^{2}\right)\right]}{(1+t)(1-t)\left(\frac{5}{4}-t\right)} \\ &1=A(1-t)\left(\frac{5}{4}-t\right)+B(1+t)\left(\frac{5}{4}-t\right)+C\left(1-t^{2}\right) \end{aligned}

\begin{aligned} &\text { Put } t=1 \\ &1=2 B\left(\frac{1}{4}\right) \\ &B=2 \end{aligned}

 

\text { Put } t=-1 \\

 

1=A(2)\left(\frac{9}{4}\right)

A=\frac{2}{9}

\begin{aligned} &\text { Put } t=\frac{5}{4} \\ &1=C\left(1-\frac{25}{16}\right) \\ &1=\frac{-9}{16} C \end{aligned}

\begin{aligned} &C=\frac{-16}{9} \\ &I=\frac{1}{4} \int \frac{\frac{2}{9}}{1+t} d t+\frac{1}{4} \int \frac{2}{(1-t)} d t-\frac{1}{4} \times \frac{16}{9} \int \frac{1}{\frac{5}{4}-t} d t \\ &I=\frac{1}{18} \log (1+t)-\frac{1}{2} \log (1-t)+\frac{4}{9} \log \left(\frac{5}{4}-t\right)+C \\ &I=\frac{1}{18} \log (1+\sin x)-\frac{1}{2} \log (1-\sin x)+\frac{4}{9} \log \left(\frac{5}{4}-\sin x\right)+C \end{aligned}

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