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Need Solution for R.D. Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 40 Maths Textbook Solution.

Answers (1)

Answer:

\frac{2(\sin x)^{\frac{3}{2}}}{3}-\frac{2}{7}(\sin x)^{\frac{7}{2}}+c

Given:

\int \sqrt{\sin x} \cos ^{3} x d x

Hint:

To solve this equation we have to suppose sin x in term of t and cos x into 2t.

Solution: 

I=\int \sqrt{\sin x} \cos ^{3} x d x                                \left[\because \sqrt{\sin x}=t, \sin x=t^{2}, \cos x d x=2 t d t\right.

    I=\int \sqrt{\sin x}\left(1-\sin ^{2} x\right) \cos x d x

    I=\int t\left(1-t^{4}\right) 2 t d t

    I=\int 2 t^{2} d t-\int 2 t^{6} d t

    I=\frac{2 t^{3}}{3}-\frac{2 t^{7}}{7}+c

    I=\frac{2(\sin x)^{\frac{3}{2}}}{3}-\frac{2}{7}(\sin x)^{\frac{7}{2}}+c

 

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