Answers 1 9 3 . \frac{1}{9} \sqrt{3}. 9 1 3 . Full solutions
(a) d d x ( cot x ) = d d x ( tan x ) − 1 = − ( tan x ) − 2 sec 2 x = − cos 2 x sin 2 x ⋅ 1 cos 2 x = − 1 sin 2 x = − cosec 2 x ■ \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \left( \cot x \right) &=
\frac{\mathrm{d}}{\mathrm{d}x} \left( \tan x \right)^{-1}
\\ &= - \left( \tan x \right)^{-2} \sec^2 x
\\ &= - \frac{\cos^2 x}{\sin^2 x} \cdot \frac{1}{\cos^2 x}
\\ &= - \frac{1}{\sin^2 x}
\\ &= - \cosec^2 x \; \blacksquare
\end{align*} d x d ( cot x ) = d x d ( tan x ) − 1 = − ( tan x ) − 2 sec 2 x = − sin 2 x cos 2 x ⋅ cos 2 x 1 = − sin 2 x 1 = − cosec 2 x ■
(b) sin 2 x tan x = ( 2 sin x cos x ) ⋅ sin x cos x = 2 sin 2 x ■ \begin{align*}
\sin 2x \tan x &= (2 \sin x \cos x ) \cdot \frac{\sin x}{\cos x}
\\ &= 2 \sin^2 x \; \blacksquare
\end{align*} sin 2 x tan x = ( 2 sin x cos x ) ⋅ cos x sin x = 2 sin 2 x ■
(c) ∫ 1 18 π 1 9 π cosec 6 x cot 3 x d x = ∫ 1 18 π 1 9 π 1 sin 6 x tan 3 x d x = ∫ 1 18 π 1 9 π 1 2 sin 2 3 x d x = 1 2 ∫ 1 18 π 1 9 π cosec 2 3 x d x = − 1 2 [ cot 3 x 3 ] 1 18 π 1 9 π = − 1 6 ( 1 tan 1 3 π − 1 tan 1 6 π ) = − 1 6 ( 1 3 − 3 ) = 1 9 3 ■ \begin{align*}
& \int_{\frac{1}{18} \pi}^{\frac{1}{9} \pi} \cosec 6x \cot 3x \mathop{\mathrm{d}x}
\\ &= \int_{\frac{1}{18} \pi}^{\frac{1}{9} \pi} \frac{1}{\sin 6x \tan 3x} \mathop{\mathrm{d}x}
\\ &= \int_{\frac{1}{18} \pi}^{\frac{1}{9} \pi} \frac{1}{2 \sin^2 3x} \mathop{\mathrm{d}x}
\\ &= \frac{1}{2} \int_{\frac{1}{18} \pi}^{\frac{1}{9} \pi} \cosec^2 3x \mathop{\mathrm{d}x}
\\ &= -\frac{1}{2} \Biggl[ \frac{\cot 3x}{3} \Biggr]_{\frac{1}{18} \pi}^{\frac{1}{9} \pi}
\\ &= -\frac{1}{6} \left( \frac{1}{\tan \frac{1}{3} \pi} - \frac{1}{\tan \frac{1}{6} \pi} \right)
\\ &= -\frac{1}{6} \left( \frac{1}{\sqrt{3}} - \sqrt{3} \right)
\\ &= \frac{1}{9} \sqrt{3} \; \blacksquare
\end{align*} ∫ 18 1 π 9 1 π cosec 6 x cot 3 x d x = ∫ 18 1 π 9 1 π sin 6 x tan 3 x 1 d x = ∫ 18 1 π 9 1 π 2 sin 2 3 x 1 d x = 2 1 ∫ 18 1 π 9 1 π cosec 2 3 x d x = − 2 1 [ 3 cot 3 x ] 18 1 π 9 1 π = − 6 1 ( tan 3 1 π 1 − tan 6 1 π 1 ) = − 6 1 ( 3 1 − 3 ) = 9 1 3 ■ Question Commentary
For part (a), if we used the relationship that co-tangent is the reciprocal of tangent,
we have to be careful not to confuse this reciprocal 1 tan x = ( tan x ) − 1 {\frac{1}{\tan x} = (\tan x)^{-1}} t a n x 1 = ( tan x ) − 1 tan − 1 x . {\tan^{-1} x.} tan − 1 x .
Solving part (c) could also be challenging as it involves using both parts (a) and (b).
Learning to pick up on the "hence" hint and being able to piece together how the two parts
can be used are vital in getting the answer.