# Evaluate the following integrals. inttan^3 4x dx

Question
Integrals
Evaluate the following integrals.
$$\int\tan^3 4x\ dx$$

2021-01-24
$$\text{To evaluate:}$$
$$\int\tan^3 4x\ dx$$
$$\tan^2 x=\sec^2 x-1$$
$$\frac{d}{dx}(\sec x)=\sec x\tan x$$
$$\text{Consider}\int\tan^3 4xdx$$
$$=\int\tan^2 4x\tan 4x dx$$
$$=\int(\sec^2 4x-1)\tan 4xdx$$
$$\text{Now, let }\sec 4x=u$$
$$\Rightarrow4\sec 4x\tan4xdx=du$$
$$\Rightarrow\tan4xdx=\frac{du}{4\sec4x}$$
$$\text{Substituting all the values, we get,}$$
$$\int (u^2-1)\frac{du}{\sec4x}$$
$$=\int(u^2-1)\frac{du}{4u}$$
$$=\frac{1}{4}\int\frac{u^2-1}{u}du$$
$$=\frac{1}{4}\int u-\frac{1}{u}du$$
$$=\frac{1}{4}[\frac{u^2}{2}-\ln(u)]+C$$
$$=\frac{1}{4}[\frac{\sec^2 4x}{2}-\ln(\sec4x)]+C$$
$$\text{Hence, }\int\tan^3 4x\ dx=\frac{1}{4}[\frac{\sec^2 4x}{2}-\ln(\sec4x)]+C$$

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