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  • Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 10

Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 10

Answers (1)

Answer: -\frac{-1}{3} \cot ^{3} x-2 \cot x+\tan x+C

Hint: - Use substitution method to solve this integration.

Given: -\int \frac{1}{\sin ^{4} x \cdot \cos ^{2} x} d x

Solution: - Let  I=\int \frac{1}{\sin ^{4} x \cdot \cos ^{2} x} d x

The power of denominator is 4+2=6so we divide both numerator and denominator by \cos ^{6}xso that the integral Ican come in integral able form.

\begin{aligned} &\therefore I=\int \frac{\frac{1}{\cos ^{6} x}}{\frac{\sin ^{4} x \cdot \cos ^{2} x}{\cos ^{6} x}} d x=\int \frac{\sec ^{6} x}{\frac{\sin ^{4} x}{\cos ^{4} x}} d x \\ &\Rightarrow=\int \frac{\sec ^{6} x}{\tan ^{4} x} d x \end{aligned}

Re-writing,

\begin{aligned} I &=\int \frac{\sec ^{4} x \cdot \sec ^{2} x}{\tan ^{4} x} d x \\ &=\int \frac{\left(\sec ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x \end{aligned}
    \begin{aligned} &=\int \frac{\left(1+\tan ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x \quad\quad\quad\quad\quad\quad\quad\quad\left[\because \sec ^{2} x-\tan ^{2} x=1\right] \\ &=\int \frac{\left(1+\tan ^{4} x+2 \tan ^{2} x\right) \sec ^{2} x}{\tan ^{4} x} \mathrm{dx} \quad\quad\quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned}    

Substitute \tan x=t \Rightarrow \sec ^{2} x d x=d t then

\begin{aligned} I &=\int \frac{\left(1+t^{4}+2 t^{2}\right)}{t^{4}} d t \\ &=\int\left(\frac{1}{t^{4}}+\frac{t^{4}}{t^{4}}+\frac{2 t^{2}}{t^{4}}\right) d t \\ &=\int\left(\frac{1}{t^{4}}+1+2 \cdot \frac{1}{t^{2}}\right) d t=\int\left(t^{-4}+t^{0}+2 t^{-2}\right) d t \end{aligned}
    \begin{aligned} &=\int\left(\frac{1}{t^{4}}+1+2 \cdot \frac{1}{t^{2}}\right) d t=\int\left(t^{-4}+t^{0}+2 t^{-2}\right) d t \\ &=\frac{t^{-4+1}}{-4+1}+\frac{t^{0+1}}{0+1}+2 \frac{t^{-2+1}}{-2+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=\frac{t^{-3}}{-3}+t+2 \frac{t^{-1}}{(-1)}+C \end{aligned}
    \begin{aligned} &=-\frac{t^{3}}{3}+t-2 t^{-1}+C=-\frac{1}{3 t^{3}}+t-\frac{2}{t}+C \\ &=-\frac{1}{3 \tan ^{3} x}+\tan x-\frac{2}{\tan x}+C \quad\quad\quad\quad\quad\quad\left [ \because t=\tan x \right ]\\ &=-\frac{1}{3} \cot ^{3} x+\tan x-2 \cot x+C\quad\quad\quad\quad\quad\quad\left [ \because \tan x=\frac{1}{\cot x} \right ] \\ &=-\frac{1}{3} \cot ^{3} x-2 \cot x+\tan x+C \end{aligned}

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