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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 69 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 69 Maths Textbook Solution.

Answers (1)

Answer:

\frac{2}{3} \tan ^{-1} \frac{\operatorname{5tan}\left(\frac{x}{2}\right)-4}{3}+c

Hint:

To solve the given statement write \sin ^{2}\left(\frac{x}{2}\right)+\cos ^{2}\left(\frac{x}{2}\right) \text { and } \sin \operatorname{as} 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)

Given:

\int \frac{1}{5-4 \sin x} d x

Solution:

I=\int \frac{1}{5-4 \sin x} d x

   =\int \frac{1}{5\left(\sin ^{2}\left(\frac{x}{2}\right)+\cos ^{2}\left(\frac{x}{2}\right)-\frac{4}{5}\left(2 \sin \frac{x}{2} \cos \left(\frac{x}{2}\right)\right)\right.} d x

   =\int \frac{d x}{5 \sin ^{2}\left(\frac{x}{2}\right)+5 \cos ^{2}\left(\frac{x}{2}\right)-8 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}

\text { Divide numerator and denominator by } \cos ^{2}\left(\frac{x}{2}\right)

     =\int \frac{\sec ^{2}\left(\frac{x}{2}\right)}{\operatorname{stan}^{2}\left(\frac{x}{2}\right)+5-8 \tan \left(\frac{x}{2}\right)} d x

     \text { let } \tan \left(\frac{x}{2}\right)=t

     \frac{\sec ^{2}\left(\frac{x}{2}\right)}{2} d x=d t

    =2 \int \frac{1}{5 t^{2}+5-8 t} d t

   =2 \times \frac{1}{5} \int \frac{1}{t^{2} \frac{(-8 t)}{5}+1} d t

   =\frac{2}{5} \int \frac{1}{t^{2}-\frac{8 t}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} d t

   =\frac{2}{5} \int \frac{d t}{\left(t-\left(\frac{4}{5}\right)^{2}\right)+\left(\frac{3}{5}\right)^{2}}

  =\frac{2}{5} \times \frac{1}{\frac{3}{5}} \tan ^{-1}\left(\frac{t-\left(\frac{4}{5}\right)}{\frac{3}{5}}\right)+c

  =\frac{2}{3} \tan ^{-1} \frac{(5 t-4)}{3}+c

  =\frac{2}{3} \tan ^{-1} \frac{\operatorname{5tan}\left(\frac{x}{2}\right)-4}{3}+c

 

Posted by

infoexpert21

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