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Question Number 46238 by Meritguide1234 last updated on 23/Oct/18

Commented by math1967 last updated on 23/Oct/18

$c o t x c o t 2 x c o t 3 x = c o t x + c o t 2 x + c o t 3 x$
Commented by Meritguide1234 last updated on 23/Oct/18

$n o$
Commented by math1967 last updated on 23/Oct/18

$c o t 3 x = \frac{c o t 2 x c o t x - 1}{c o t 2 x + c o t x}$ $∴ c o t 3 x c o t 2 x + c o t 3 x c o t x = c o t 2 x c o t x - 1$ $s o t h a t w a s e r r o r m y c o m m e n t$
Commented by MJS last updated on 23/Oct/18

$why not try these :$ $5 . \int \sec x \sec 2 x \sec 3 x d x$ $6 . \int \csc x \csc 2 x \csc 3 x d x$
Commented by Meritguide1234 last updated on 23/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

	$3) t a n 3 x = t a n (x + 2 x)$ $= \frac{t a n x + t a n 2 x}{1 - t a n x . t a n 2 x}$ $t a n 3 x - t a n x t a n 2 x t a n 3 x = t a n x + t a n 2 x$ $t a n 3 x - t a n x - t a n 2 x = t a n x t a n 2 x t a n 3 x$ $s o \int t a n x t a n 2 x t a n 3 x d x$ $= \int (t a n 3 x - t a n 2 x - t a n x) d x$ $= \frac{1}{3} \times l n s e c 3 x - \frac{1}{2} \times l n s e c 2 x - l n s e c x + c$ $= c o r r e c t e d$
	Commented by Meritguide1234 last updated on 23/Oct/18

	$l a s t l i n e i n t e g r a n d i s$ $\frac{1}{3} l n s e c 3 x - \frac{1}{2} l n s e c 2 x - l n s e c x + c$
	Commented by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

	$y e s i n h u r r y . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

	$1) \int s i n x s i n 2 x s i n 3 x d x$ $= \frac{1}{2} \int s i n x (2 s i n 2 x s i n 3 x) d x$ $= \frac{1}{2} \int s i n x (c o s x - c o s 5 x) d x$ $= \frac{1}{4} \int {(2 s i n x c o s x) - (2 s i n x c o s 5 x)} d x$ $= \frac{1}{4} \int {(s i n 2 x - (s i n 6 x - s i n 4 x)} d x$ $= \frac{1}{4} \int (s i n 2 x - s i n 6 x + s i n 4 x) d x$ $= \frac{1}{4} {\frac{- c o s 2 x}{2} + \frac{c o s 6 x}{6} + \frac{- c o s 4 x}{4}} + c$
	Commented by Meritguide1234 last updated on 23/Oct/18

	$c o r r e c t$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18
	$2)(1/2)∫cosx(2cos2xcos3x)dx =(1/2)∫cosx(cos5x+cosx)dx =(1/4)∫{2cosxcos5x+2cos^2 x}dx =(1/4)∫{cos6x+cos4x+cos2x+1}dx (1/4){((sin6x)/6)+((sin4x)/4)+((sin2x)/2)+x}+c$
	$2) \frac{1}{2} \int c o s x (2 c o s 2 x c o s 3 x) d x$ $= \frac{1}{2} \int c o s x (c o s 5 x + c o s x) d x$ $= \frac{1}{4} \int {2 c o s x c o s 5 x + 2 {c o s}^{2} x} d x$ $= \frac{1}{4} \int {c o s 6 x + c o s 4 x + c o s 2 x + 1} d x$ $\frac{1}{4} {\frac{s i n 6 x}{6} + \frac{s i n 4 x}{4} + \frac{s i n 2 x}{2} + x} + c$
	Commented by Meritguide1234 last updated on 23/Oct/18

	$c o r r e c t . . . t r y n o 4$
	Commented by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

	$y e s s i r i a m t r y i n g Q 4 . . .$
	Answered by MJS last updated on 23/Oct/18

	$\int \cot x \cot 2 x \cot 3 x d x =$ $= \int (\cot x \times \frac{\cos^{2} x - \sin^{2} x}{2 \cos x \sin x} \times \frac{\cos x (\cos^{2} x - {3 \sin}^{2} x)}{\sin x ({3 \cos}^{2} x - \sin^{2} x}) d x =$ $[\cos x = \frac{\cot x}{\csc x}; \sin x = \frac{1}{\csc x}; \csc^{2} x = 1 + \cot^{2} x]$ $= \int \frac{\csc^{2} x \cot x (\cot^{4} x - {4 \cot}^{2} x + 3)}{2 ({3 \cot}^{4} x + {2 \cot}^{2} x - 1)} d x =$ $[t = \cot x \to d x = - \frac{d t}{\csc^{2} x}]$ $= - \frac{1}{2} \int \frac{t (t^{4} - 4 t^{2} + 3)}{3 t^{4} + 2 t^{2} - 1} d t = - \frac{1}{6} \int t d t + \frac{1}{3} \int \frac{t (7 t^{2} - 5)}{3 t^{4} + 2 t^{2} - 1} d t$ $- \frac{1}{6} \int t d t = - \frac{t^{2}}{12} = - \frac{\cot^{2} x}{12}$ $\frac{1}{3} \int \frac{t (7 t^{2} - 5)}{3 t^{4} + 2 t^{2} - 1} d t = \int \frac{t}{t^{2} + 1} d t - \frac{2}{3} \int \frac{t}{3 t^{2} - 1} d t$ $\int \frac{t}{t^{2} + 1} d t = \frac{1}{2} \ln (t^{2} + 1) = \frac{1}{2} \ln (1 + \cot^{2} x) = \ln ∣ \csc x ∣$ $- \frac{2}{3} \int \frac{t}{3 t^{2} - 1} d t = - \frac{1}{9} \ln (3 t^{2} - 1) = - \frac{1}{9} \ln ∣ 1 - {3 \cot}^{2} x ∣$ $\int \cot x \cot 2 x \cot 3 x d x =$ $= \ln ∣ \csc x ∣ - \frac{1}{9} \ln ∣ 1 - {3 \cot}^{2} x ∣ - \frac{\cot^{2} x}{12} + C$
	Commented by Meritguide1234 last updated on 23/Oct/18

	$g o o d a p p r o a c h$
	Commented by Meritguide1234 last updated on 23/Oct/18

	Commented by Meritguide1234 last updated on 23/Oct/18

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