Solved: "Integral Calculus: Trigonometry and Inverse Trigonometry. Evaluate the..."

James Dale

Answered question

2021-12-10

Integral Calculus: Trigonometry and Inverse Trigonometry. Evaluate the following.

\int \frac{\sin θ (\cos θ + 4) d θ}{1 + \cos^{2} θ}

Answer & Explanation

lalilulelo2k3eq

Beginner2021-12-11Added 38 answers

Step 1
Let,
$I = \int \frac{\sin θ (\cos θ + 4) d θ}{1 + \cos^{2} θ}$
Step 2
$\int \frac{\sin θ (\cos θ + 4) d θ}{1 + \cos^{2} θ}$
$= \int \frac{\sin θ \cos θ d θ}{1 + \cos^{2} θ} + \int \frac{4 \sin θ d θ}{1 + \cos^{2} θ}$
$= I_{1} + I_{2}$
Now,
$I_{1} = \int \frac{\sin θ (\cos θ + 4) d θ}{1 + \cos^{2} θ}$
Put $1 + \cos^{2} θ = t \Rightarrow 2 \cos θ (- \sin θ) d θ = d t \Rightarrow \cos θ \sin θ = \frac{- 1}{2} d t$
$∴ I_{1} = \int \frac{- 1 d t}{2 t}$
$= \frac{- 1}{2} \ln (t) + c$
$= \frac{\ln (1 + \cos^{2} θ)}{2} + c$
Now,
$I_{2} = \int \frac{4 \sin θ d θ}{1 + \cos^{2} θ}$
Put
$= - 4 \tan^{- 1} (t) + d$
$= - 4 \tan^{{- 1}} (\cos θ) + d$
Thus,
$I = I_{1} + I_{2}$
$= \frac{- \ln (1 + \cos^{2} θ)}{2} + c - 4 \tan^{{- 1}} (\cos θ) + d$
$= \frac{- \ln (1 + \cos^{2} θ)}{2} - 4 \tan^{{- 1}} (\cos θ) + C$

Serita Dewitt

Beginner2021-12-12Added 41 answers

Perhaps if not for you, not for your help in mathematics, I would not have been able to get credit, thank you so much for that !!!

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