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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.

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


\int \tan ^{4} x d x


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


\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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