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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.13 Question 2

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.13 Question 2

Answers (1)

Answer: \frac{1}{8a^{2}}\frac{x^{8}}{\left ( a^{2}-x^{2} \right )^{4}}+c

Hint: Use substitution method to solve this integral.

Given: \int \frac{x^{7}}{\left ( a^{2}-x^{2} \right )^{5}} dx

Solution:

Let I=\int \frac{x^{7}}{\left ( a^{2}-x^{2} \right )^{5}} dx

Put x=a \sin\theta\Rightarrow dx=a\cos\theta d\theta           …(i)                     (Differentiating w.r.t to θ)

\begin{aligned} &\text { Then } \mathrm{I}=\int \frac{a^{7} \sin ^{7} \theta}{\left(a^{2}-a^{2} \sin ^{2} \theta\right)^{5}} \cdot a \cos \theta d \theta \\ &=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{\left\{a^{2}\left(1-\sin ^{2} \theta\right)\right\}^{5}} d \theta=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{\left.a^{2 \times 5}\left(\cos ^{2} \theta\right)\right\}^{5}} d \theta \\ &=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{a^{10} \cos ^{10} \theta} d \theta=\frac{1}{a^{2}} \int \frac{\sin ^{7} \theta}{\cos ^{9} \theta} d \theta \end{aligned}

\begin{aligned} &=\frac{1}{a^{2}} \int \frac{\sin ^{7} \theta}{\cos ^{7} \theta} \cdot \frac{1}{\cos ^{2} \theta} d \theta \\ &=\frac{1}{a^{2}} \int \tan ^{7} \theta \cdot \sec ^{2} \theta d \theta \quad \quad\quad\quad\quad\quad\quad\left[\because \frac{\sin \theta}{\cos \theta}=\tan \theta \text { and } \frac{1}{\cos \theta}=\sec \theta\right] \end{aligned}
Again put\tan\theta=t\Rightarrow \sec2\theta d\theta=dt          (on differentiating w.r.t to \theta)

\begin{aligned} &\text { Then } I=\frac{1}{a^{2}} \int t^{7} d t=\frac{1}{a^{2}} \cdot \frac{t^{7+1}}{7+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &I=\frac{1}{a^{2}} \cdot \frac{t^{8}}{8}+c=\frac{1}{8 a^{2}} \cdot t^{8}+c \\ &I=\frac{1}{8 a^{2}} \cdot t^{8}+c \quad\quad\quad\quad\quad[\because t=\tan \theta] \end{aligned}

From (i)x=a\sin\theta

\Rightarrow \sin \theta=\frac{x}{a}

and we know that \sin^{2}\theta+\cos^{2}\theta=1

\begin{aligned} &\Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \\ &\Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-\left(\frac{x}{a}\right)^{2}} \\ &\Rightarrow \cos \theta=\sqrt{1-\frac{x^{2}}{a^{2}}}=\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}=\frac{1}{a} \sqrt{a^{2}-x^{2}} \end{aligned}

\begin{aligned} &\therefore \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{x}{a}}{\frac{1}{a} \sqrt{a^{2}-x^{2}}} \\ &\Rightarrow \tan \theta=\frac{x}{a} \cdot \frac{a}{\sqrt{a^{2}-x^{2}}}=\frac{x}{\sqrt{a^{2}-x^{2}}} \end{aligned}

 

\therefore \tan\theta=\frac{x}{\sqrt{a^{2}-x^{2}}}                                             …(iii)

 

Put the value of tanθ from (iii) in (ii) then

\begin{aligned} &I=\frac{1}{8 a^{2}}\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right)^{8}+c \\ &=\frac{1}{8 a^{2}} \frac{x^{8}}{\left(a^{2}-x^{2}\right)^{\frac{8}{2}}}+c \\ &=\frac{1}{8 a^{2}} \frac{x^{8}}{\left(a^{2}-x^{2}\right)^{4}}+c \end{aligned}

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