Evaluate the following integrals. inttan^3 4x dx

Evaluate the following integrals. inttan^3 4x dx

Question
Integrals
asked 2021-01-23
Evaluate the following integrals.
\(\int\tan^3 4x\ dx\)

Answers (1)

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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