Get the answer to "Evaluate the integral. ∫sec60tan20d0"

Phoebe

Answered question

2021-11-05

Evaluate the integral.

\int \frac{\sec^{6} 0}{\tan^{2} 0} d 0

Macsen Nixon

Skilled2021-11-06Added 117 answers

Step 1
Consider the following integral:

I = \int \frac{\sec^{6} 0}{\tan^{2} 0} d 0

= \int \sec^{2} (0) \cdot \frac{\sec^{4} (0)}{\tan^{2} (0)} d 0

= \int \sec^{2} (0) \cdot \frac{{[\sec^{2} (0)]}^{2}}{\tan^{2} (0)} d 0

= \int \sec^{2} (0) \cdot \frac{[1 + \tan^{2} (0)]}{\tan^{2} (0)} d 0

Step 2
Substitute

\tan (0) = t \to \sec^{2} 0 d 0 = d t

in I:

I = \int \frac{{(t^{2} + 1)}^{2}}{t^{2}} d t

= \int \frac{t^{4} + 2 t^{2} + 1}{t^{2}} d t

= \int (t^{2} + 2 + \frac{1}{t^{2}}) d t

= \int t^{2} d t + \int 2 d t + \int t^{- 2} d t

= \frac{t^{3}}{3} + 2 t - \frac{1}{t} + C

Step 3
Substitute

t = \tan (0)

in I:

I = \frac{\tan^{3} (0)}{3} + 2 \tan (0) - \frac{1}{\tan (0)} + C

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