Step by step solution to "Evaluate the following integrals. ∫sin3θcos−2θdθ"

amanf

Answered question

2021-10-05

Evaluate the following integrals.

\int \sin^{3} θ \cos^{- 2} θ d θ

Theodore Schwartz

Skilled2021-10-06Added 99 answers

Let the given integral be I, then:

I = \int \sin^{3} θ \cos^{- 2} θ d θ

Rewrite the above equation, using trigonometric formula

\sin^{2} θ + \cos^{2} θ = 1

I = \int \frac{\sin^{3} θ}{\cos^{2} θ} d θ

= \int \frac{(\sin^{2} θ) \sin θ}{\cos^{2} θ} d θ

= \int \frac{(1 - \cos^{θ}) \sin θ}{\cos^{2} θ} d θ

(1)
Take:

t = \cos θ

(2)
Differentiate t w.r.t \theta:

\frac{d t}{d θ} = \frac{d}{d θ} (\cos θ)

= - \sin θ

- d t = \sin θ d θ

(3)
Substitute the value of (2) and (3) in (1):

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

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

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

Solve the integral:

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

= t + \frac{1}{t} + C

Where C is a constant.
Now, from (2) substitute the value of t in above equation:

I = \cos θ + \frac{1}{\cos θ} + C

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