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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 130 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 130 Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{3} \log |\cot x+1|-\frac{1}{6} \log \left|\cot ^{2} x-\cot x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cot x-1}{\sqrt{3}}\right)+C

Given: \int \frac{\cot x+\cot ^{3} x}{1+\cot ^{3} x} d x

Hint: using partial fraction and \int \frac{1}{x} d x, \int \frac{1}{1+x^{2}} d x

Explanation:

\begin{aligned} &I=\int \frac{\cot x+\cot ^{3} x}{1+\cot ^{3} x} d x \\ &=\int \frac{\cot x\left(1+\cot ^{2} x\right)}{1+\cot ^{3} x} d x \\ &=\int \frac{\cot x \cos e c^{2} x}{1+\cot ^{3} x} d x \\ &=-\int \frac{t}{1+t^{3}} d t \end{aligned}                                                   \left[\begin{array}{l} p u t \cot x=t \\ -\cos e c^{2} x=d t \\ \cos e c^{2} x d x=-d t \end{array}\right]

=-\int \frac{t}{(1+t)\left(t^{2}-t+1\right)}

\begin{aligned} &\text { now: } \frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{A}{1+t}+\frac{B}{t^{2}-t+1} \\ &\text { multiplying by }(1+t)\left(t^{2}-t+1\right) \\ &t=A\left(t^{2}-t+1\right)+(B t+C)(t+1) \end{aligned}

\begin{aligned} &\text { putting\: t }=-1 \\ &-1=A(1+1+1)+(B(-1)+C)(0) \\ &=3 A \Rightarrow A=-\frac{1}{3} \end{aligned}

\begin{aligned} &\text { putting } t=0 \\ &0=A(0-0+1)+(B(0)+C)(0+1) \\ &0=A(1)+C(1) \\ &0=A+C \Rightarrow 0=\frac{-1}{3}+C \Rightarrow C=\frac{1}{3} \end{aligned}

\begin{aligned} &\text { putting } t=+1 \\ &1=A(1-1+1)+(B+C)(2) \\ &1=A(1)+2 B+2 C \\ &1=\frac{-1}{3}+2 B+\frac{2}{3} \end{aligned}

\begin{aligned} &1+\frac{1}{3}-\frac{2}{3}=2 B \\ &2 B=\frac{3+1-2}{3} \\ &2 B=\frac{2}{3} \Rightarrow B=\frac{1}{3} \end{aligned}

\frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{\frac{-1}{3}}{t+1}+\frac{\frac{1}{3}(t+1)}{t^{2}-t+1}

\int \frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{3} \int \frac{t+1}{t^{2}-t+1} d t

=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{3 \times 2} \int \frac{2 t+2-3+3}{t^{2}-t+1} d t

=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{6} \int \frac{2 t-1}{t^{2}-t+1} d t-\frac{1}{2} \int \frac{1}{t^{2}-t+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1}

=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{6} \int \frac{2 t-1}{t^{2}-t+1} d t-\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}}

=\frac{1}{3} \log |t+1|-\frac{1}{6} \log \left|t^{2}-t+1\right|-\frac{1}{2} \cdot \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C

=\frac{1}{3} \log |t+1|-\frac{1}{6} \log \left|t^{2}-t+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C

=\frac{1}{3} \log |\cot x+1|-\frac{1}{6} \log \left|\cot ^{2} x-\cot x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cot x-1}{\sqrt{3}}\right)+C

 

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