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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 54 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 54 Maths Textbook Solution.

Answers (2)

Answer: -\frac{3}{2} \sqrt{x} \cos \sqrt{x}+\frac{3}{2} \sin \sqrt{x}+\frac{1}{6}\left[\sqrt{x} \cos 3 \sqrt{x}-\frac{\sin 3 \sqrt{x}}{3}\right]+c

Hint: Put \sqrt{x}=t

Given: \int \sin ^{3}\sqrt{xdx}

Solution: Put

                   \sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}

                   \Rightarrow dx=2tdt

            \begin{aligned} &\therefore \int \sin ^{3} \sqrt{x} d x=2 \int t \sin ^{3} t d t \\ &{\left[\begin{array}{l} \therefore \sin 3 t=3 \sin t-4 \sin ^{3} t \\ \sin ^{3} t=\frac{3 \sin t-\sin 3 t}{4} \end{array}\right]} \\ &\Rightarrow 2 \int \frac{t[3 \sin t-\sin 3 t]}{4} d t \\ &\Rightarrow \frac{3}{2} \int t \sin t d t-\frac{1}{2} \int t \sin 3 t d t \end{aligned}

Applying by parts,

                \begin{aligned} &=\frac{3}{2}\left[t(-\cos t)-\int(-\cos t) d t\right]-\frac{1}{2}\left[t\left(\frac{-\cos 3 t}{3}\right)-\int\left(\frac{-\cos 3 t}{3}\right) d t\right] \\ &=\frac{3}{2}[-t \cos t+\sin t]-\frac{1}{2}\left[\frac{-t \cos 3 t}{3}+\frac{1}{3} \frac{\sin 3 t}{3}\right]+c \\ &=-\frac{3}{2} \sqrt{x} \cos \sqrt{x}+\frac{3}{2} \sin \sqrt{x}+\frac{1}{6}\left[\sqrt{x} \cos 3 \sqrt{x}-\frac{\sin 3 \sqrt{x}}{3}\right]+c \end{aligned}

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Answer: -\frac{3}{2} \sqrt{x} \cos \sqrt{x}+\frac{3}{2} \sin \sqrt{x}+\frac{1}{6}\left[\sqrt{x} \cos 3 \sqrt{x}-\frac{\sin 3 \sqrt{x}}{3}\right]+c

Hint: Put \sqrt{x}=t

Given: \int \sin ^{3}\sqrt{xdx}

Solution: Put

                   \sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}

                   \Rightarrow dx=2tdt

            \begin{aligned} &\therefore \int \sin ^{3} \sqrt{x} d x=2 \int t \sin ^{3} t d t \\ &{\left[\begin{array}{l} \therefore \sin 3 t=3 \sin t-4 \sin ^{3} t \\ \sin ^{3} t=\frac{3 \sin t-\sin 3 t}{4} \end{array}\right]} \\ &\Rightarrow 2 \int \frac{t[3 \sin t-\sin 3 t]}{4} d t \\ &\Rightarrow \frac{3}{2} \int t \sin t d t-\frac{1}{2} \int t \sin 3 t d t \end{aligned}

Applying by parts,

                \begin{aligned} &=\frac{3}{2}\left[t(-\cos t)-\int(-\cos t) d t\right]-\frac{1}{2}\left[t\left(\frac{-\cos 3 t}{3}\right)-\int\left(\frac{-\cos 3 t}{3}\right) d t\right] \\ &=\frac{3}{2}[-t \cos t+\sin t]-\frac{1}{2}\left[\frac{-t \cos 3 t}{3}+\frac{1}{3} \frac{\sin 3 t}{3}\right]+c \\ &=-\frac{3}{2} \sqrt{x} \cos \sqrt{x}+\frac{3}{2} \sin \sqrt{x}+\frac{1}{6}\left[\sqrt{x} \cos 3 \sqrt{x}-\frac{\sin 3 \sqrt{x}}{3}\right]+c \end{aligned}

Posted by

infoexpert21

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