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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 14 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 14 maths Textbook Solution.

Answers (1)

Answer: 10 \ln |\sin x-4|-7 \ln |\sin x-3|+c

Hint : Let \sin x=t

Given: \int \frac{(3 \sin x-2) \cos x}{13-\cos ^{2} x-7 \sin x} d x

Solution: \text { Let } \sin x=t

                        \cos x d x=d t

                      \int \frac{(3 \sin x-2) \cos x}{13-\left(1-\sin ^{2} x\right)-7 \sin x} d x                                                             \left(\cos ^{2} x=1-\sin ^{2} x\right)

            \int \frac{(3 t-2) d t}{13-1+t^{2}-7 t}=\int \frac{(3 t-2) d t}{t^{2}-7 t+12}                                                              \left[\begin{array}{l} t^{2}-7 t+12=(t-4)(t-3) \\ a x^{2}+b x+c, x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \end{array}\right]

  \int \frac{3 t-2}{(t-4)(t-3)} d t=\frac{A}{t-4}+\frac{B}{t-3}                                            ..........(1)

\begin{aligned} &3 t-2=\frac{A}{t-4}+\frac{B}{t-3} \\ &3 t-2=A(t-3)+B(t-4) \\ &3 t-2=A t-3 A+B t-4 B \end{aligned}

On comparing,

3 \mathrm{t}=\mathrm{At}+\mathrm{Bt} \Rightarrow \mathrm{A}+\mathrm{B}=3 \Rightarrow>\mathrm{A}=3-\mathrm{B}

-2=-3 A-4 B=23 A+4 B=2=>3(3-B)+4 B=2=>B=-7

On solving both
A=3-(-7) \Rightarrow A=10

In eqn (1)

            \begin{aligned} & \int\left[\frac{10}{t-4}+\frac{-7}{t-3}\right] d t \\ =& \int \frac{10}{t-4} d t-\int \frac{7}{t-3} d t \end{aligned}                                                                    \int \frac{1}{t} d t=\ln |t|+c

   \begin{aligned} &10 \ln |t-4|-7 l n|t-3|+c \\ &=10 \ln |t-4|-7 \ln |t-3|+c \\ &=10 \ln |\sin x-4|-7 \ln |\sin x-3|+c \end{aligned}

Posted by

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