What is the Integral of ${tan}^{3} 3 x \cdot sec 3 x d x$ $tan^3 3x * sec3x dx$ ?

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What is the Integral of ${tan}^{3} 3 x \cdot sec 3 x d x$ $tan^3 3x * sec3x dx$ ?

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Answer 1 · 2018-05-24T17:10:53

$\int {tan}^{3} 3 x sec 3 x d x = \frac{1}{9} {sec}^{3} 3 x - \frac{1}{3} sec 3 x + c$ $inttan^3 3xsec3x"d"x=1/9sec^3 3x-1/3sec3x+"c"$

Explanation:

For $\int {tan}^{3} 3 x sec 3 x d x$ $inttan^3 3xsec3x"d"x$ , let $u = 3 x$ $u=3x$ and $d u = 3 d x$ $"d"u=3"d"x$

Then $\int {tan}^{3} 3 x sec 3 x d x = \frac{1}{3} \int {tan}^{3} u sec u d u$ $inttan^3 3xsec3x"d"x=1/3inttan^3usecu"d"u$

Now, use ${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x=sec^2x-1$

$\frac{1}{3} \int {tan}^{3} u sec u d u = \frac{1}{3} \int ({sec}^{2} u - 1) tan u sec u d u = \frac{1}{3} \int {sec}^{3} u tan u d u - \frac{1}{3} \int sec u tan u d u$ $1/3inttan^3usecu"d"u=1/3int(sec^2u-1)tanusecu"d"u=1/3intsec^3utanu"d"u-1/3intsecutanu"d"u$

For the first integral, we use the reverse chain rule, noting that $\frac{d}{d x} \frac{1}{3} {sec}^{3} u = {sec}^{3} u tan u$ $d/dx1/3sec^3u=sec^3utanu$ and for the second, we use the fundamental theorem of calculus. So,

$\frac{1}{3} \int {sec}^{3} u tan u d u - \frac{1}{3} \int sec u tan u d u = \frac{1}{9} {sec}^{3} u - \frac{1}{3} sec u$ $1/3intsec^3utanu"d"u-1/3intsecutanu"d"u=1/9sec^3u-1/3secu$

Now we substitute $u = 3 x$ $u=3x$ to get a final answer of

$\frac{1}{9} {sec}^{3} 3 x - \frac{1}{3} sec 3 x + c$ $1/9sec^3 3x-1/3sec3x+"c"$

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What is the Integral of ${tan}^{3} 3 x \cdot sec 3 x d x$ $tan^3 3x * sec3x dx$ ?

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