Expert answers to "Evaluate the following integrals. ∫tan3⁡4x dx"

defazajx

Answered question

2021-01-23

Evaluate the following integrals.

\int \tan^{3} 4 x d x

broliY

Skilled2021-01-24Added 97 answers

To evaluate:

\int \tan^{3} 4 x d x

\tan^{2} x = \sec^{2} x - 1

\frac{d}{d x} (\sec x) = \sec x \tan x

Consider \int \tan^{3} 4 x d x

= \int \tan^{2} 4 x \tan 4 x d x

= \int (\sec^{2} 4 x - 1) \tan 4 x d x

Now, let \sec 4 x = u

\Rightarrow 4 \sec 4 x \tan 4 x d x = d u

\Rightarrow \tan 4 x d x = \frac{d u}{4 \sec 4 x}

Substituting all the values, we get,

\int (u^{2} - 1) \frac{d u}{\sec 4 x}

= \int (u^{2} - 1) \frac{d u}{4 u}

= \frac{1}{4} \int \frac{u^{2} - 1}{u} d u

= \frac{1}{4} \int u - \frac{1}{u} d u

= \frac{1}{4} [\frac{u^{2}}{2} - \ln (u)] + C

= \frac{1}{4} [\frac{\sec^{2} 4 x}{2} - \ln (\sec 4 x)] + C

Hence, \int \tan^{3} 4 x d x = \frac{1}{4} [\frac{\sec^{2} 4 x}{2} - \ln (\sec 4 x)] + C

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