Find: 1. ∫ cos3x cosx dx = ? 2. ∫ 3^x sinx dx = ? 3. ∫_(0 ) ^( 1) x e^(−x) dx = ? 4. ∫


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Question Number 201763 by hardmath last updated on 11/Dec/23

$Find :$ $1 . \int \cos 3 x cosx dx = ?$ $2 . \int 3^{x} sinx dx = ?$ $3 . \int_{0}^{1} x e^{- x} dx = ?$ $4 . \int_{1}^{e} \ln^{2} x dx = ?$
	Answered by Calculusboy last updated on 12/Dec/23

	$S o l u t i o n :$ $(3) \int_{0}^{1} {x e}^{- x} d x (b y u s i n g {B e r n o u l l i}^{'} s f o r m u l a)$ $u = x u^{'} = 1 u^{″} = 0 a n d v = e^{- x} v_{1} = - e^{- x} v_{2} = e^{- x}$ $v_{3} = - e^{- x}$ $I = {u v}_{1} - u^{'} v_{2} + u^{″} v_{3} - u^{‴} v_{4} + \cdot \cdot \cdot$ $I = - {x e}^{- x} - e^{- x}$ $I = - {[{x e}^{- x}]}_{0}^{1} - {[e^{- x}]}_{0}^{1}$ $I = - [e^{- 1} - 0] - [e^{- 1} - 1]$ $I = - e^{- 1} - e^{- 1} + 1$ $I = 1 - \frac{2}{e} = \frac{e - 2}{e} ✓$
	Commented by hardmath last updated on 12/Dec/23

	$thank you ser$
	Answered by Calculusboy last updated on 12/Dec/23
	$Solution: (2) ∫ 3^x sinx dx (by using IBP) let u=3^x du=3^x In(3)dx dv=sinx v=−cosx ∫udvdx=uv−∫vdudx let I=∫udvdx I=−3^x cosx+In(3)∫3^x cosx dx let I_1 =∫3^x cosxdx let u=3^x du=3^x In(3)dx dv=cosx v=sinx I_1 =uv−∫vdudx I_1 =3^x sinx−In(3)∫3^x sinxdx but I=∫3^x sinxdx I_1 =3^x sinx−In(3)I and I=−3^x cosx+In(3)I_1 I=−3^x cosx+In(3)[3^x sinx−In(3)I] I=−3^x cosx+3^x In(3)sinx−In^2 (3)I I+In^2 (3)I=3^x In(3)sinx−3^x cosx I=(3^x /( [1+In^2 (3)])){In(3)sinx−cosx}+C$
	$S o l u t i o n :$ $(2) \int 3^{x} s i n x d x (b y u s i n g I B P)$ $l e t u = 3^{x} d u = 3^{x} I n (3) d x d v = s i n x v = - c o s x$ $\int u d v d x = u v - \int v d u d x l e t I = \int u d v d x$ $I = - 3^{x} c o s x + I n (3) \int 3^{x} c o s x d x$ $l e t I_{1} = \int 3^{x} c o s x d x l e t u = 3^{x} d u = 3^{x} I n (3) d x d v = c o s x v = s i n x$ $I_{1} = u v - \int v d u d x$ $I_{1} = 3^{x} s i n x - I n (3) \int 3^{x} s i n x d x b u t I = \int 3^{x} s i n x d x$ $I_{1} = 3^{x} s i n x - I n (3) I a n d I = - 3^{x} c o s x + I n (3) I_{1}$ $I = - 3^{x} c o s x + I n (3) [3^{x} s i n x - I n (3) I]$ $I = - 3^{x} c o s x + 3^{x} I n (3) s i n x - {I n}^{2} (3) I$ $I + {I n}^{2} (3) I = 3^{x} I n (3) s i n x - 3^{x} c o s x$ $I = \frac{3^{x}}{[1 + {I n}^{2} (3)]} {I n (3) s i n x - c o s x} + C$
	Commented by hardmath last updated on 12/Dec/23

	$thank you ser,$ $I n = \ln . ?$
	Commented by Calculusboy last updated on 12/Dec/23

	$i m e a n i t l i k e t h i s L n$
	Commented by mr W last updated on 12/Dec/23

	$y o u k n o w t h e d i f f e r e n b e t w e e n I a n d$ $l, b u t y o u w o n t m a k e t h e d i f f e r e n c e$ $a n d t h e r e f o r e c a u s e c o n f u s i o n s$ $a g a i n a n d a g a i n .$ $s e e Q 201546$
	Commented by hardmath last updated on 12/Dec/23

	$thank you dear professors$
	Commented by Calculusboy last updated on 12/Dec/23

	$a m s o r r y f o r t h e c o n f u s i n g s i r$
	Answered by Calculusboy last updated on 12/Dec/23

	$S o l u t i o n :$ $(1) \int c o s 3 x c o s x d x$ $R e c a l l t h a t \int c o s a x c o s b x d x = \frac{1}{2} \int [c o s (a - b) x + c o s (a + b) x] d x$ $I = \frac{1}{2} \int [c o s (3 - 1) x + c o s (3 + 1) x] d x$ $I = \frac{1}{2} \int [c o s 2 x + c o s 4 x] d x$ $I = \frac{1}{2} [\int c o s 2 x d x + \int c o s 4 x d x]$ $I = \frac{1}{2} [\frac{s i n 2 x}{2} + \frac{s i n 4 x}{4}] + C$
	Answered by esmaeil last updated on 12/Dec/23

	$4 : = Ω l n x d x = d v \to v = x l n x - x$ $l n x = u \to \frac{d x}{x} = d u$ $\to Ω = {x l n}^{2} x - x l n x - ({\int_{1}}^{e} (l n x - 1) d x) =$ $Missing \left or extra \right$ $e - 2$
	Answered by Calculusboy last updated on 12/Dec/23

	$S o l u t i o n :$ $(4) \int_{1}^{e} {L n}^{2} x d x = \int_{1}^{e} {(L n x)}^{2} d x (b y u s i n g u - s u b)$ $l e t p = L n x x d p = d x x = e^{p}$ $w h e n x = e p = 1 a n d w h e n x = 1 p = 0$ $I = \int_{0}^{1} p^{2} \cdot e^{p} d p (b y u s i n g B F m e t h o d)$ $u = p^{2} u^{'} = 2 p u^{″} = 2 u^{‴} = 0$ $v = e^{p} v_{1} = e^{p} v_{2} = e^{p} v_{3} = e^{p} v_{4} = e^{p}$ $I = {u v}_{1} - u^{'} v_{2} + u^{″} v_{3} - u^{‴} v_{4} + \cdot \cdot \cdot$ $I =∣ p^{2} e^{p} ∣_{0}^{1} - 2 ∣ {p e}^{p} ∣_{0}^{1} + 2 ∣ e^{p} ∣_{0}^{1} + C$ $I = (e - 0) - 2 (e - 0) + 2 (e - 1)$ $I = e - 2 e + 2 e - 2$ $I = e - 2$
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