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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.13 Question 5

Answers (1)

Answer: \frac{1}{54}\left[\tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{3(x+1)}{x^{2}+2+10}\right]+c

Hint: Use substitution method to solve this integral

Given:\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x \\


\begin{aligned} &\text { Let } I=\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x \\ &=\int \frac{1}{\left(x^{2}+2 x+1+9\right)^{2}} d x \\ &=\int \frac{1}{\left\{\left(x^{2}+2 x \cdot 1+1\right)+9\right\}^{2}} d x \\ &=\int \frac{1}{\left\{(x+1)^{2}+9\right\}^{2}} d x \end{aligned}


Putting x+1=3\tan\theta                                      …(i)


d x=3 \sec ^{2} \theta d \theta


\begin{aligned} &I=\int \frac{1}{\left\{(3 \tan \theta)^{2}+9\right\}^{2}} 3 \sec ^{2} \theta d \theta \\ &=\int \frac{3 \sec ^{2} \theta}{\left\{9\left(1+\tan ^{2} \theta\right)\right\}^{2}} d \theta=\int \frac{3 \sec ^{2} \theta}{9^{2}\left(\sec ^{2} \theta\right)^{2}} d \theta \end{aligned}


\begin{aligned} &=\int \frac{3 \sec ^{2} \theta}{81 \sec ^{4} \theta} d \theta \quad\left[\begin{array}{l} \because \sec ^{2} \theta-\tan ^{2} \theta=1 \\ \Rightarrow 1+\tan ^{2} \theta=\sec ^{2} \theta \end{array}\right] \\ &=\int \frac{1}{27} \frac{1}{\sec ^{2} \theta} d \theta=\int \frac{1}{27} \cos ^{2} \theta d \theta \quad\left[\because \cos \theta=\frac{1}{\sec \theta}\right] \\ &=\frac{1}{27} \int\left\{\frac{1+\cos 2 \theta}{2}\right\} d \theta \quad\left[\begin{array}{l} \because 2 \cos ^{2} A=1+\cos 2 A \\ \cos ^{2} A=\frac{1+\cos 2 A}{2} \end{array}\right] \end{aligned}


\begin{aligned} &=\frac{1}{27}\left\{\int \frac{1}{2} d \theta+\int \frac{1}{2} \cos 2 \theta d \theta\right\} \\ &=\frac{1}{27} \times \frac{1}{2} \int \theta^{0} d \theta+\frac{1}{27} \times \frac{1}{2} \int \cos 2 \theta d \theta \\ &=\frac{1}{54} \cdot \frac{\theta^{0+1}}{0+1}+\frac{1}{54} \frac{\sin 2 \theta}{2}+c\quad\quad\quad\quad[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \text { and } \int \cos a x d x=\frac{\sin a x}{a}+c ]\\ \end{aligned}

=\frac{1}{54} \theta+\frac{1}{108} \sin 2 \theta+c                                   …(ii)

Also from (i)

\begin{aligned} &x+1=3 \tan \theta \\ &\tan \theta=\frac{x+1}{3} \Rightarrow \theta=\tan ^{-1}\left(\frac{x+1}{3}\right) \end{aligned} … (iii)


\begin{array}{ll} \Rightarrow \tan ^{2} \theta=\left(\frac{x+1}{3}\right)^{2} & \text { [Squaring both sides] } \\\\ \Rightarrow \frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{(x+1)^{2}}{9} \quad & {\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]} \\ \\\Rightarrow \sin ^{2} \theta=\frac{(x+1)^{2}}{9} \cdot \cos ^{2} \theta & {\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]} \end{array}

\begin{aligned} &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right) \cos ^{2} \theta}{9} \\ &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right)}{9}\left(1-\sin ^{2} \theta\right) \quad\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \end{array}\right] \\ &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right)}{9}-\left(\frac{\left(x^{2}+2 x+1\right)}{9}\right) \sin ^{2} \theta \end{aligned}

\begin{aligned} &\Rightarrow \frac{x^{2}+2 x+1}{9}=\sin ^{2} \theta+\left(\frac{x^{2}+2 x+1}{9}\right) \sin ^{2} \theta \\ &\Rightarrow \frac{x^{2}+2 x+1}{9}=\sin ^{2} \theta\left\{1+\left(\frac{x^{2}+2 x+1}{9}\right)\right\} \\ &\Rightarrow \frac{x^{2}+2 x+1}{9}=\left(\frac{9+x^{2}+2 x+1}{9}\right) \sin ^{2} \theta=\left(\frac{x^{2}+2 x+10}{9}\right) \sin ^{2} \theta \end{aligned}

\begin{aligned} &\Rightarrow \sin ^{2} \theta=\frac{x^{2}+2 x+1}{9} \times \frac{9}{x^{2}+2 x+10}=\frac{(x+1)^{2}}{\left(x^{2}+2 x+10\right)} \quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\ &\Rightarrow \sin \theta=\sqrt{\frac{(x+1)^{2}}{x^{2}+2 x+10}}=\frac{(x+1)}{\sqrt{x^{2}+2 x+10}} \quad \ldots \text { (iv) } \\ &\quad \text { Since } \sin ^{2} \theta+\cos ^{2} \theta=1 \end{aligned}


 So we can write


\begin{aligned} &\Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \\ &\Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta} \\ &\begin{aligned} \Rightarrow \cos \theta &=\sqrt{1-\left[\frac{(x+1)}{\sqrt{x^{2}+2 x+10}}\right]^{2}} \\ &=\sqrt{\frac{x^{2}+2 x+10-\left(x^{2}+2 x+1\right)}{x^{2}+2 x+10}} \end{aligned} \quad\quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned}

\begin{aligned} &\Rightarrow \cos \theta=\sqrt{\frac{x^{2}+2 x+10-x^{2}-2 x-1}{x^{2}+2 x+10}}\\ &\Rightarrow \cos \theta=\sqrt{\frac{9}{x^{2}+2 x+10}}=\frac{3}{\sqrt{x^{2}+2 x+10}}\quad\quad\quad\quad....(v) \end{aligned}


Also, we know \sin 2 \theta=2 \sin \theta \cos \theta


\begin{aligned} &\Rightarrow \sin 2 \theta=2 \frac{(x+1)}{\sqrt{x^{2}+2 x+10}} \cdot \frac{3}{\sqrt{x^{2}+2 x+10}} \quad\quad \quad \quad [\text { from (iv) and (v)] }\\ &\Rightarrow \sin 2 \theta=\frac{2.3(x+1)}{x^{2}+2 x+10}=\frac{6(x+1)}{x^{2}+2 x+10}\quad \quad \quad \quad \cdot \cdot \cdot \cdot (vi) \end{aligned}

Now put the value of θ and sin2θ from equation (iii) and (vi) in (ii) we get
\begin{aligned} &I=\frac{1}{54} \theta+\frac{1}{108} \sin 2 \theta+c \\ &=\frac{1}{54} \tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{1}{108} \cdot \frac{6(x+1)}{\left(x^{2}+2 x+10\right)}+c \\ &=\frac{1}{54} \tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{1}{54} \frac{3(x+1)}{\left(x^{2}+2 x+10\right)}+c \\ &=\frac{1}{54}\left[\tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{3(x+1)}{\left(x^{2}+2 x+10\right)}\right]+c \end{aligned}


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