explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 72 maths

Answer: $\frac{1}{3 \cos ^{2} x}-\cos x-\frac{2}{\cos x}+c$

Hint: Use substitution method to solve this integral

Given: $\int \frac{\sin ^{5} x}{\cos ^{4} x} d x$

Solution:

\begin{aligned} &\text { Let } I=\int \frac{\sin ^{5} x}{\cos ^{4} x} d x \\ &\text { Put } \cos \mathrm{x}=t \Rightarrow-\sin x \; d x=d t \\ &\Rightarrow d x=\frac{d t}{-\sin x} \text { then } \end{aligned}

$I=\int \frac{\sin ^{5} x}{t^{4} } \frac{d t}{-\sin x}=-\int \frac{\sin ^{4} x}{t^{4}} d t$

$=-\int \frac{\left(\sin ^{2} x\right)^{2}}{t^{4}} d t=-\int \frac{\left(1-\cos ^{2} x\right)^{2}}{t^{4}} d t$

$I=-\int \frac{\left(1+\cos ^{4} x-2 \cos ^{2} x\right)}{t^{4}} d t \quad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

$=-\int\left[\frac{1+t^{4}-2 t^{2}}{t^{4}}\right] d t \quad[\because \cos x=t]$

\begin{aligned} &=-\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-\frac{2}{t^{2}}\right] d t \end{aligned}

\begin{aligned} &=-\int\left(t^{-4}+1-2 t^{-2}\right) d t \\ &=-\int t^{-4} d t-\int t^{0} d t+2 \int t^{-2} d t \end{aligned}

$=-\frac{t^{-4+1}}{-4+1}-\frac{t^{0+1}}{0+1}+2 \frac{t^{-2+1}}{-2+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

\begin{aligned} &=-\frac{t^{-3}}{-3}-t+2 \frac{t^{-1}}{-1}+c \\ &=\frac{1}{3 t^{3}}-t-2 \cdot \frac{t^{-1}}{1}+c \end{aligned}

$=\frac{1}{3 \cos ^{2} x}-\cos x-\frac{2}{\cos x}+c \quad[\because t=\cos x]$