Evaluate the integral. \int \tan^{5}xdx

Question

Evaluate the integral. \int \tan^{5}xdx

Holly Guerrero 2021-12-31 Answered

Evaluate the integral.

\int \tan^{5} x d x

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Expert Answer

Alex Sheppard

Answered 2022-01-01 Author has 36 answers

Step 1
Split

\tan^{5} x

into simpler powers

\tan^{3} x

and

\tan^{2} x

and change

\tan^{2} x

to

\sec^{2} x - 1

and simplify.

\int \tan^{5} x d x = \int \tan^{3} x (\tan^{2} x d x)

= \int \tan^{3} x (\sec^{2} x - 1) d x

= \int (\tan^{3} x \sec^{2} x - \tan^{3} x) d x

Step 2
use the integral theorem and integrate both the of

\int (\tan^{3} x \sec^{2} x - \tan^{3} x) d x

terms with respect to dx and simplify to obtain the value of the integral.

\int (\tan^{3} x \sec^{2} x - \tan^{3} x) d x = \int \tan^{3} x \sec^{2} x d x - \int \tan^{3} x d x

= \int (\tan^{3} x \sec^{2} x) d x - \int \tan^{2} x d x (\tan x) d x

= \int \tan^{3} x \sec^{2} x d x - \int (\sec^{2} x - 1) \tan x d x

= \int \tan^{3} x \sec^{2} x d x - \int \tan x \sec^{2} x d x + \int \tan x d x

= \frac{1}{4} \tan^{4} x - \frac{1}{2} \tan^{2} x + \ln \sec x + C

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Linda Birchfield

Answered 2022-01-02 Author has 39 answers

\int {\tan (x)}^{5} d x

Evaluate the integral

\frac{1}{4} \times {\tan (x)}^{4} - \int {\tan (x)}^{3} d x

Evaluate the integral

\frac{1}{4} \times {\tan (x)}^{4} - (\frac{1}{2} \times {\tan (x)}^{2} - \int \tan (x) d x)

Expand the expression

\frac{1}{4} \times {\tan (x)}^{4} - (\frac{1}{2} \times {\tan (x)}^{2} - \int \frac{\sin (x)}{\cos (x)} d x)

Transform the expression

\frac{1}{4} \times {\tan (x)}^{4} - (\frac{1}{2} \times {\tan (x)}^{2} - \int - \frac{1}{t} d t)

Use properties of integrals

\frac{1}{4} \times {\tan (x)}^{4} - (\frac{1}{2} \times {\tan (x)}^{2} + \int \frac{1}{t} d t)

Evaluate the integral

\frac{1}{4} \times {\tan (x)}^{4} - (\frac{1}{2} \times {\tan (x)}^{2} + \ln (| t |))

Substitute back

\frac{1}{4} \times {\tan (x)}^{4} - (\frac{1}{2} \times {\tan (x)}^{2} + \ln (| \cos (x) |))

Simplify

\frac{{\tan (x)}^{4}}{4} - \frac{{\tan (x)}^{2}}{2} - \ln (| \cos (x) |)

Add C
Solution

\frac{{\tan (x)}^{4}}{4} - \frac{{\tan (x)}^{2}}{2} - \ln (| \cos (x) |) + C

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karton

Answered 2022-01-04 Author has 368 answers

$\begin{matrix} \tan^{5} (x) d x \\ = \int (\tan^{2} (x))^{2} \tan (x) d x \\ = \int (\sec^{2} (x) - 1)^{2} \tan (x) d x \\ = \int \frac{(u^{2} - 1)^{2}}{u} d u \\ = \int (u^{3} - 2 u + \frac{1}{u}) d u \\ = \int u^{3} d u - 2 \int u d u + \int \frac{1}{u} d u \\ \int u^{3} d u \\ = \frac{u^{4}}{4} \\ \int u d u \\ = \frac{u^{2}}{2} \\ \int \frac{1}{u} d u \\ = \ln (u) \\ \int u^{3} d u - 2 \int u d u + \int \frac{1}{u} d u \\ = \ln (u) + \frac{u^{4}}{4} - u^{2} \\ = \ln (\sec (x)) + \frac{\sec^{4} (x)}{4} - \sec^{2} (x) \\ \int \tan^{5} (x) d x \\ = \ln (| \sec (x) |) + \frac{\sec^{4} (x)}{4} - \sec^{2} (x) + C \end{matrix}$