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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 37 Maths Textbook Solution.

Answers (1)

Answer:

\frac{\cos ^{5} x}{5}-\frac{\cos ^{7} x}{7}+C

Given:

\int \sin ^{3} x \cos ^{4} x d x

Hint:

To solve the statement we will convert cos into sin and use some formula of 2sin(A+B) and 2sin(A-B)

Solution:

\int \sin ^{3} x\cos ^{4}x d x

put

\int \sin x \cdot \sin ^{2} x \cdot \cos ^{4} x d x

p u t \: \cos x=t, \sin x d x=d t

  \int \sin x\left(1-\cos ^{2} x\right) \cdot \cos ^{4} x d x\left[\sin ^{2} x=1-\cos ^{2} x\right]

   \int\left(1-t^{2}\right) t^{4} d t

      \int t^{4}-t^{6} d t

    \int t^{4} d t-\int t^{6} d t

 \int \frac{t^{5}}{5}-\frac{t^{7}}{7}

   t=\cos x

\frac{\cos ^{5} x}{5}-\frac{\cos ^{7} x}{7}+C

 

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infoexpert21

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