Find the integrals of the functions in Exercises: ^4x - Vidyadip

Question

Find the integrals of the functions in Exercises:
$\tan^4\text{x}$

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Answer

$\tan^4\text{x}$
$=\tan^2\text{x}\tan^2\text{x}$
$=\big(\sec^2\text{x}-1\big)\tan^2\text{x}$
$=\sec^2\text{x}\tan^2\text{x}-\tan^2\text{x}$
$=\sec^2\text{x}\tan^2\text{x}-\big(\sec^2\text{x}-1\big)$
$=\sec^2\text{x }\tan^2\text{x}-\sec^2\text{x}+1$
$\therefore\int\tan^4\text{x}\text{ dx}=\int\sec^2\text{x }\tan^2\text{x}\text{ dx}-\int\sec^2\text{x}\text{ dx}+\int1\cdot\text{dx}$
$=\int\sec^2\text{x }\tan^2\text{x}\text{ dx}-\tan\text{x}+\text{x}+\text{C} \ \ \ \ ...\text{(1)}$
Consider $\int\sec^2\text{x }\tan^2\text{x}\text{ dx}$
$\text{Let }\tan\text{x}=\text{t}\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\int\sec^2\text{x }\tan^2\text{x }\text{dx}=\int\text{t}^2\text{dt}=\frac{\text{t}^3}{3}=\frac{\tan^3\text{x}}{3}$
From equation(1),we obtain
$\int\tan^4\text{x}\text{ dx}=\frac{1}{3}\tan^3\text{x}-\tan\text{x}+\text{x}+\text{C}$

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