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Integration Using Substitution

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Question

Find the general solutions of the following equations:
(i) $\sin 2 θ = \frac{\sqrt{3}}{2}$
(ii) $\cos 3 θ = \frac{1}{2}$
(iii) $\sin 9 θ = \sin θ$
(iv) $\sin 2 θ = \cos 3 θ$
(v) $\tan θ + \cot 2 θ = 0$
(vi) $\tan 3 θ = \cot θ$
(vii) $\tan 2 θ \tan θ = 1$
(viii) $\tan m θ + c o t n θ = 0$
(ix) $\tan p θ = c o t q θ$
(x) $\sin 2 θ + \cos θ = 0$
(xi) $\sin θ = \tan θ$
(xii) $\sin 3 θ + \cos 2 θ = 0$

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Solution

We have:

(i) $\sin 2 θ = \frac{\sqrt{3}}{2}$
⇒ $\sin 2 θ = \sin \frac{π}{3}$
⇒ $2 θ = n π + (- 1)^{n} \frac{π}{3}$ $n \in Z$
⇒ $θ = \frac{n π}{2} + (- 1)^{n} \frac{π}{6}$ , $n \in Z$

(ii) $\cos 3 θ = \frac{1}{2}$
⇒ $\cos 3 θ = \cos \frac{π}{3}$
⇒ $3 θ = 2 n π \pm \frac{π}{3}$ $n \in Z$
⇒ $θ = \frac{2 n π}{3} \pm \frac{π}{9}$ , $n \in Z$

(iii) $\sin 9 θ = \sin θ$
⇒ $\sin 9 θ - \sin θ = 0$
⇒ $2 \sin (\frac{9 θ - θ}{2}) \cos (\frac{9 θ + θ}{2}) = 0$
⇒ $\sin \frac{8 θ}{2} = 0$ or $\cos \frac{10 θ}{2} = 0$
⇒ $\sin 4 θ = 0$ or $\cos 5 θ = 0$
⇒ $4 θ = n π$ , $n \in Z$ or $5 θ = (2 n + 1) \frac{π}{2}$ , $n \in Z$
⇒ $θ = \frac{n π}{4}$ , $n \in Z$ or $θ = (2 n + 1) \frac{π}{10}$ , $n \in Z$

(iv) $\sin 2 θ = \cos 3 θ$
$\cos 3 θ = \sin 2 θ$
⇒ $\cos 3 θ = \cos (\frac{π}{2} - 2 θ)$
⇒ $3 θ = 2 n π \pm (\frac{π}{2} - 2 θ), n \in Z$
On taking positive sign, we have:
$3 θ = 2 n π + (\frac{π}{2} - 2 θ)$
⇒ $5 θ = 2 n π + \frac{π}{2}$
⇒ $θ = \frac{2 n π}{5} + \frac{π}{10}$
⇒ $θ = (4 n + 1) \frac{π}{10}$ , $n \in Z$

Now, on taking negative sign, we have:
$3 θ = 2 n π - \frac{π}{2} + 2 θ, n \in Z$
⇒ $θ = 2 n π - \frac{π}{2}$

⇒ $θ = (4 n - 1) \frac{π}{2}, n \in Z$

(v) $\tan θ + \cot 2 θ = 0$

$\Rightarrow \tan θ = - \cot 2 θ \Rightarrow \tan θ = \tan (\frac{π}{2} + 2 θ) \Rightarrow θ = n π + (\frac{π}{2} + 2 θ), n \in Z \Rightarrow - θ = n π + \frac{π}{2}, n \in Z \Rightarrow θ = - n π - \frac{π}{2}, n \in Z \Rightarrow θ = mπ - \frac{π}{2}, m = - n \in Z$

(vi) $\tan 3 θ = c o t θ$
$\Rightarrow \tan 3 θ = \tan (\frac{π}{2} - θ) \Rightarrow 3 θ = n π + (\frac{π}{2} - θ), n \in Z \Rightarrow 4 θ = n π + \frac{π}{2}, n \in Z \Rightarrow θ = \frac{n π}{4} + \frac{π}{8}, n \in Z$

(vii) $\tan 2 θ \tan θ = 1$
$\Rightarrow \tan 2 θ = \frac{1}{\tan θ} \Rightarrow \tan 2 θ = c o t θ \Rightarrow \tan 2 θ = \tan (\frac{π}{2} - θ) \Rightarrow 2 θ = n π + (\frac{π}{2} - θ), n \in Z \Rightarrow 3 θ = n π + \frac{π}{2}, n \in Z \Rightarrow θ = \frac{n π}{3} + \frac{π}{6}, n \in Z$

(viii) $\tan m θ + c o t n θ = 0$

$\Rightarrow \tan m θ = - \cot n θ \Rightarrow \tan m θ = \tan (\frac{π}{2} + n θ) \Rightarrow m θ = r π + (\frac{π}{2} + n θ), r \in Z \Rightarrow (m - n) θ = r π + \frac{π}{2}, r \in Z \Rightarrow θ = (\frac{2 r + 1}{m - n}) \frac{π}{2}, r \in Z$

(ix) $\tan p θ = c o t q θ$
$\Rightarrow \tan p θ = \tan (\frac{π}{2} - q θ) \Rightarrow p θ = n π + (\frac{π}{2} - q θ), n \in Z \Rightarrow (p + q) θ = n π + \frac{π}{2}, n \in Z \Rightarrow θ = (\frac{2 n + 1}{p + q}) \frac{π}{2}, n \in Z$

(x) $\sin 2 θ + \cos θ = 0$
$\Rightarrow \cos θ = - \sin 2 θ \Rightarrow \cos θ = \cos (\frac{π}{2} + 2 θ) \Rightarrow θ = 2 n π \pm (\frac{π}{2} + 2 θ), n \in Z$
On taking positive sign, we have:
$θ = 2 n π + \frac{π}{2} + 2 θ \Rightarrow - θ = 2 n π + \frac{π}{2} \Rightarrow θ = 2 m π - \frac{π}{2}, m = - n \in Z \Rightarrow θ = \frac{(4 m - 1) π}{2}, m \in Z$
On taking negative sign, we have:
$θ = 2 n π - \frac{π}{2} - 2 θ \Rightarrow 3 θ = 2 n π - \frac{π}{2} \Rightarrow θ = \frac{(4 n - 1) π}{6}, n \in Z$

(xi) $\sin θ = \tan θ$
$\Rightarrow \sin θ - \tan θ = 0 \Rightarrow \sin θ - \frac{\sin θ}{\cos θ} = 0 \Rightarrow \sin θ (1 - \frac{1}{\cos θ}) = 0 \Rightarrow \sin θ (\cos θ - 1) = 0$

$\Rightarrow \sin θ = 0$ or $\cos θ - 1 = 0$
Now,
$\sin θ = 0 \Rightarrow θ = n π, n \in Z$

$\cos θ - 1 = 0 \Rightarrow \cos θ = 1 \Rightarrow \cos θ = \cos 0 \Rightarrow θ = 2 m π, m \in Z$

(xii) $\sin 3 θ + \cos 2 θ = 0$
$\Rightarrow \cos 2 θ = - \sin 3 θ \Rightarrow \cos 2 θ = \cos (\frac{π}{2} + 3 θ) \Rightarrow 2 θ = 2 n π \pm (\frac{π}{2} + 3 θ), n \in Z$
On taking positive sign, we have:
$2 θ = 2 n π + \frac{π}{2} + 3 θ \Rightarrow - θ = 2 n π + \frac{π}{2} \Rightarrow θ = 2 m π - \frac{π}{2}, m = - n \in Z \Rightarrow θ = \frac{(4 m - 1) π}{2}, m \in Z$
On taking negative sign, we have:
$2 θ = 2 n π - \frac{π}{2} - 3 θ \Rightarrow 5 θ = 2 n π - \frac{π}{2} \Rightarrow θ = \frac{(4 n - 1) π}{10}, n \in Z$

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\sin 2 x = \frac{\sqrt{3}}{2}

\cos 3 x = \frac{1}{2}

\sin 9 x = \sin x

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θ

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ} = \frac{{sec}^{2} θ}{{tan}^{2} θ - 1}

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