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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 29 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 29 Maths Textbook Solution.

Answers (1)

Answer:

\frac{\tan ^{8} x}{3}-\tan x+x+c

Given:

\int \tan ^{4} x d x

Hint:

To solve this statement we have to change tan into sec and then their formula.

Solution:

\int \tan ^{4} x d x

    I=\int \tan ^{2} x \cdot \tan ^{2} x \cdot d x \quad\left[\because \tan ^{2} x=\sec ^{2} x-1\right]

\mathrm{I}=\int \tan ^{2} x\left(\sec ^{2} x-1\right) d x

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

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

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

 \tan x=t, \sec ^{2} d x=d t

 \int t^{2} d t-\int \sec ^{2} x d x+\int d x

=\frac{t^{8}}{3}-\tan x+x+c

\frac{\tan ^{8} x}{3}-\tan \mathrm{x}+\mathrm{x}+\mathrm{c}

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