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  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 56 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 56  Maths Textbook Solution.

Answers (1)

Answer: =2 \sqrt{x} \sin \sqrt{x}+\frac{4}{3} \cos \sqrt{x}-\frac{2 \sqrt{x} \sin ^{3} \sqrt{x}}{3}+\frac{2}{9}\left[\cos ^{3} \sqrt{x}\right]+c

Hint: Let \sqrt{x}=t

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

Solution:  \sqrt{x}=t\Rightarrow x=t^{2}\Rightarrow dx=2tdt

                \begin{aligned} &\therefore \int \cos ^{3} \sqrt{x} d x=\int \cos ^{3} t .2 t d t \Rightarrow 2 \int t \cos ^{3} t d t \\ &=2 \int t \cos t \cos ^{2} t d t \Rightarrow 2 \int t \cdot \cos t\left(1-\sin ^{2} t\right) d t \\ &=2 \int\left(t \cos t-t \cos t \sin ^{2} t\right) d t \\ &=2\left[\int t \cos t d t-\int t \cos t \sin ^{2} t d t\right] \\ &=2\left[t \int \cos t d t-\int[1 \cdot \sin t d t]\right]-2\left[t \int \cos \sin ^{2} t d t-\int 1 \cdot\left(\frac{\sin ^{3} t}{3}\right) d t\right] \end{aligned}

Applying by parts

\begin{aligned} &=2\left(t \sin t-\int \sin t d t\right)-2\left(t \cdot \frac{\sin ^{3} t}{3}-\frac{1}{3} \int \sin ^{3} t d t\right) \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3} \int \sin t \cdot \sin ^{2} t d t \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3} \int \sin t \cdot\left(1-\cos ^{2} t\right) d t \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3}\left[\int \sin t d t-\int \sin t \cos ^{2} t d t\right] \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3}\left[-\cos t+\frac{\cos ^{3} t}{3}+c\right] \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3}[-\cos t]+\frac{2}{3}\left[\frac{\cos ^{3} t}{3}\right]+c \\ &=2 \sqrt{x} \sin \sqrt{x}+\frac{4}{3} \cos \sqrt{x}-\frac{2 \sqrt{x} \sin ^{3} \sqrt{x}}{3}+\frac{2}{9}\left[\cos ^{3} \sqrt{x}\right]+c \end{aligned}

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