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 \(\mathrm{m}+\mathrm{n}=\)
- A 8
- B 12
- C 10
- D 16
Answer & Solution
Correct Answer
(B) 12
Step-by-step Solution
Detailed explanation
Let \(I=\int \sec ^4 \tan ^4 x d x\)
Put \(\tan \mathrm{x}=\mathrm{t} \Rightarrow \sec ^2 \mathrm{x} d \mathrm{x}=\mathrm{dt}\)
\(\therefore \mathrm{I}=\int \sec ^2 \mathrm{x}\left(\sec ^2 \mathrm{x}\right)\left(\tan ^4 \mathrm{x}\right) \mathrm{dx} \)
\( =\int\left(1+\mathrm{t}^2\right)(\mathrm{t})^4 \mathrm{dt}=\int\left(\mathrm{t}^4+\mathrm{t}^6\right) \mathrm{dt}=\frac{\mathrm{t}^5}{5}\) \(+\frac{\mathrm{t}^7}{7}+\mathrm{c}=\frac{\tan ^5 \mathrm{x}}{5}+\frac{\tan ^7 \mathrm{x}}{7}+\mathrm{c}\)
Put \(\tan \mathrm{x}=\mathrm{t} \Rightarrow \sec ^2 \mathrm{x} d \mathrm{x}=\mathrm{dt}\)
\(\therefore \mathrm{I}=\int \sec ^2 \mathrm{x}\left(\sec ^2 \mathrm{x}\right)\left(\tan ^4 \mathrm{x}\right) \mathrm{dx} \)
\( =\int\left(1+\mathrm{t}^2\right)(\mathrm{t})^4 \mathrm{dt}=\int\left(\mathrm{t}^4+\mathrm{t}^6\right) \mathrm{dt}=\frac{\mathrm{t}^5}{5}\) \(+\frac{\mathrm{t}^7}{7}+\mathrm{c}=\frac{\tan ^5 \mathrm{x}}{5}+\frac{\tan ^7 \mathrm{x}}{7}+\mathrm{c}\)
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