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

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

Answers (1)

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]

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