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MHT CET · Maths · Indefinite Integration

\(\int \sec ^4 x \cdot \tan ^4 x d x=\frac{\tan ^m x}{m}+\frac{\tan ^n x}{n}+c\) (where \(c\) is constant of integration), then \(\mathrm{m}+\mathrm{n}=\)

  1. A 8
  2. B 12
  3. C 10
  4. D 16
Verified Solution

Answer & Solution

Correct Answer

(B) 12

Step-by-step Solution

Detailed explanation

Let \(I=\int \sec ^4 \tan ^4 x d x\)
Put \(\tan \mathrm{x}=\mathrm{t} \Rightarrow \sec ^2 \mathrm{x} d \mathrm{x}=\mathrm{dt}\)
\(\therefore \mathrm{I}=\int \sec ^2 \mathrm{x}\left(\sec ^2 \mathrm{x}\right)\left(\tan ^4 \mathrm{x}\right) \mathrm{dx} \)
\( =\int\left(1+\mathrm{t}^2\right)(\mathrm{t})^4 \mathrm{dt}=\int\left(\mathrm{t}^4+\mathrm{t}^6\right) \mathrm{dt}=\frac{\mathrm{t}^5}{5}\) \(+\frac{\mathrm{t}^7}{7}+\mathrm{c}=\frac{\tan ^5 \mathrm{x}}{5}+\frac{\tan ^7 \mathrm{x}}{7}+\mathrm{c}\)