Question
Write a value of $\int\tan^3\text{x}\sec^2\text{x}\text{ dx}$
✓
Answer
Let $\text{I}=\int\tan^3\text{x}\sec^2\text{x}\text{ dx}$ Let $\tan\text{x}=\text{t}$ $\sec^2\text{x dx}=\text{dt}$ $\therefore\ \text{I}=\int\text{t}^3\text{ dt}$ $=\frac{\text{t}^4}{4}+\text{C}$$=\frac{\tan^{4}\text{x}}{4}+\text{C}$ $(\because\text{t}=\tan\text{x})$
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