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  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 38 Maths Textbbok Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise  Revision Exercise Question 38 Maths Textbbok Solution.

Answers (1)

Answer:

\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x-\cos x+c

Given:

\int \sin ^{5} x d x

Hint:

To solve the statement we have to convert sin x into cos x.

Solution: 

\int \sin x(\sin x)^{4} d x

   \int \sin x\left(\sin ^{2} x\right)^{2} d x                            \left[\because \sin ^{2} \theta+\cos ^{2} \theta=1, \sin ^{2} \theta=1-\cos ^{2} \theta\right]

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

 Let cosx= t

-\sin x d x=d t

\sin x d x=-d t

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

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

I=\int\left(2 t^{2}-t^{4}-1\right) d t \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right]

I=\frac{2 t^{3}}{3}-\frac{t^{5}}{5}-t+c

I=\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x-\cos x+c

 

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