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  • Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.24 question 1 sub question (xi) maths textbook solution

Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.24 question 1 sub question (xi) maths textbook solution

 

Answers (1)

Answer: I=\frac{9}{41} \log |4 \operatorname{Cos} x+5 \sin x|+\frac{40 x}{41}+C

Hint: use the formula in which

Put Numerator = λ denominator+ μ (derivative of denominator)

Given: I=\int \frac{4 \sin x+5 \cos x}{5 \sin x+4 \cos x} \mathrm{~d} \mathrm{x}

Explanation:

\text { let, } I=\int \frac{4 \sin x+5 \cos x}{5 \sin x+4 \cos x} \mathrm{~d} \mathrm{x}

So, we will take the steps as described

4 \sin x+5 \cos x=A(5 \cos x-4 \sin x)+B(4 \cos x+5 \sin x)

        \left(\frac{d}{d x} \operatorname{cos} x=-\sin x\right)

4 \sin x+5 \cos x=(5 B-4 A) \sin x+(5 A+4 B) \cos x

Comparing both sides, we have

\begin{aligned} &5 \mathrm{~B}-4 \mathrm{~A}=4 \quad \ldots (i)\\ &4 B+5 A=5 \quad \ldots(ii) \end{aligned}

Multiplying (i) by 5 and (ii) by 4 and then add;

25 B-20 A+16 B+20 A=20+20

On solving for A and B , we have

        A=\frac{9}{41}, B=\frac{40}{41}

Thus, I can be expressed as

        \mathrm{I}=\int \frac{\frac{9}{41}(5 \cos x-4 \sin x)+\frac{40}{41}(4 \cos x+5 \sin x)}{4 \cos x+5 \sin x} d x

        I=\int \frac{9}{41}\left(\frac{5 \cos x-4 \sin x}{4 \cos x+5 \sin x}\right) \mathrm{d} \mathrm{x}+\int \frac{40}{41}\left(\frac{4 \cos x+5 \sin x}{4 \cos x+5 \sin x}\right) \mathrm{dx}

\begin{aligned} &\text { Let, } I_{1}=\frac{9}{41} \int\left(\frac{5 \cos x-4 \sin x}{4 \cos x+5 \sin x}\right) \mathrm{d} \mathrm{x} \\ &\mathrm{I}_{2}=\frac{40}{41} \int\left(\frac{4 \cos x+5 \sin x}{4 \cos x+5 \sin x}\right) \mathrm{d} \mathrm{x} \end{aligned}

        \begin{array}{ll} I=I_{1}+I_{2} & \ldots \text { equ. }(iii) \end{array}

        I_{1}=\frac{9}{41} \int\left(\frac{5 \cos x-4 \sin x}{4 \cos x+5 \sin x}\right) d x

\begin{aligned} &\text { Let } 4 \cos x+5 \sin x=u \\ &\Rightarrow(-4 \sin x+5 \cos x) d x=d u \end{aligned}

So, I_{1}  reduces to

        I_{1}=\frac{9}{41} \int \frac{d u}{u}=\frac{9}{41} \log |u|+c_{1}

Therefore,

        I_{1}=\frac{9}{41} \log |4 \operatorname{Cos} x+5 \sin x|+c_{1} \quad \ldots \text { Eq. (iv) }

\text { As, }I_{2}=\frac{40}{41} \int \frac{4 \cos x+5 \sin x}{4 \cos x+5 \sin x} \mathrm{dx}

        I_{2}=\frac{40}{41} \int d x=\frac{40}{41} \times+c_{2} \quad \text { ... Eq. }(v)

From Equ (iii) , (iv) , (v), we have

        I=\frac{9}{41} \log |4 \cos x+5 \sin x|+c_{1}+\frac{40}{41} x+c_{2}

Therefore,

        I=\frac{9}{41} \log |4 \cos x+5 \sin x|+\frac{40}{41} x+c \quad\left[c_{1}+c_{2}=c\right]

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