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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 39 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 39 Maths Textbook Solution.

Answers (1)

Answer:

\sin x+\frac{\sin ^{5} x}{5}-\frac{2 \sin ^{8} x}{3}+c

Given:

\int \cos ^{5} x d x

Hint:

To solve the statement we have to convert cos term in sin

Solution:

\int \cos ^{4} x \cos x d x

  I=\left(\cos ^{2} x\right)^{2} \cos x d x \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right]

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

\sin x=t, \cos 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=t+\frac{t^{5}}{5}-\frac{2 t^{3}}{3}+c

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

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