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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 30 Maths Textbook Solution.

Answers (1)

Answer:

\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c

Given:

\int \tan ^{5} x d x

Hint:

To solve this statement we have to change tan into sec and then use formula such tan²x.

Solution: 

I=\int \tan ^{5} x d x

  I=\int \tan ^{3} x \tan ^{2} x d x

      =\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x \quad\left[\because \tan ^{2} x=\sec ^{2} x-1\right]

      =\int \tan ^{3} x \sec ^{2} x d x-\int \tan ^{3} x d x

      =\int \tan ^{3} x \sec ^{2} x d x-\int \tan x \tan ^{2} x d x

                       I_{1}                                    I_{2}

   \operatorname{let} \tan x=t, \sec ^{2} x d x=d t

    =\int t^{3} d t-\int \tan x \tan ^{2} x d x

    =\frac{t^{4}}{4}-\int \tan x\left(\sec ^{2} x-1\right) d x

     =\frac{t^{4}}{4}-\int \tan x \sec ^{2} x d x+\int \tan x d x

    =\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c

     =\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c

  

 

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