Check the solution to "Evaluate the following. ∫sin⁡θ(cos⁡θ+4)dθ1+cos2θ"

foass77W

Answered question

2021-02-09

Evaluate the following.

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

gotovub

Skilled2021-02-10Added 98 answers

Let,

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

I = \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 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

\cos θ = t \Rightarrow - \sin θ d θ = d t \Rightarrow \sin θ d θ = - d t

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

= - 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

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