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Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 2

Answers (1)

Answer:  -\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+C

Hint: Use substitution method to solve this integral

Given: \int \sin^{5 }xdx

Solution: Let, I=\int \sin^{5 }xdx


\begin{aligned} &I=\int \sin ^{3} x \sin ^{2} x d x \\ &=\int \sin ^{3} x\left(1-\cos ^{2} x\right) \quad \quad \quad \quad \quad \quad\left[\because \sin ^{2} x+\cos ^{2} x=1\right] \end{aligned}    
 \begin{aligned} &=\int\left(\sin ^{3} x-\sin ^{3} x \cos ^{2} x\right) d x \\ &=\int\left(\sin ^{2} x \sin x-\sin ^{3} x \cos ^{2} x\right) d x \\ &\left.=\int\left(1-\cos ^{2} x\right) \sin x-\sin ^{3} x \cos ^{2} x\right) d x \quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x=1-\cos ^{2} x\right] \end{aligned}
 \begin{aligned} &=\int\left(\sin x-\sin x \cos ^{2} x-\sin ^{3} x \cos ^{2} x\right) d x \\ &=\int \sin x d x-\int \sin x \cos ^{2} x-\int \sin ^{3} x \cos ^{2} x d x \end{aligned}

Substitute \cos x=t \Rightarrow-\sin x d x \Rightarrow d t in second and third integral, then                                    

\begin{aligned} I &=\int \sin x d x-\int t^{2}(-d t)-\int t^{2} \sin ^{3} x \frac{d t}{-\sin x} \\ &=\int \sin x d x+\int t^{2} d t+\int \sin ^{2} x t^{2} d t \\ &=\int \sin x d x+\int t^{2} d t+\int\left(1-\cos ^{2} x\right) t^{2} d t\quad\quad\quad\quad\quad\quad&\left[\because \sin ^{2} x=1-\cos ^{2} x\right] \\ \end{aligned}
    \begin{aligned} &=\int \sin x d x+\int t^{2} d t+\int t^{2}\left(1-t^{2}\right) d t \\ &=\int \sin x d x+\int t^{2} d t+\int t^{2}-t^{2} \cdot t^{2} d t \\ &=\int \sin x d x+\int t^{2} d t+\int t^{2} d t-\int t^{4} d t \end{aligned}
    \begin{aligned} &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{4+1}}{4+1}+C \quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \sin x d x=-\cos x\right] \\ &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{5}}{5}+C \end{aligned}
    \begin{aligned} &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{t^{3}}{3}+\frac{t^{3}}{3}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{2 t^{3}}{3}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5}+C \quad \quad \quad \quad \quad \quad \quad[\because t=\cos x] \end{aligned}

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