Math Question and Answers


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 121887 by oustmuchiya@gmail.com last updated on 12/Nov/20

Commented by liberty last updated on 12/Nov/20

$(vi) {(2 \cos x - \sin x)}^{2} = 4 \cos^{2} x - 4 \cos xsin x + \sin^{2} x$ $= 4 (\frac{1}{2} + \frac{1}{2} \cos 2 x) - 2 \sin 2 x + \frac{1}{2} - \frac{1}{2} \cos 2 x$ $= 2 + 2 \cos 2 x - 2 \sin 2 x + \frac{1}{2} - \frac{1}{2} \cos 2 x$ $= \frac{5}{2} + \frac{3}{2} \cos 2 x - 2 \sin 2 x$ $I = \int (\frac{5}{2} + \frac{3}{2} \cos 2 x - 2 \sin 2 x) dx$ $I = \frac{5 x}{2} + \frac{3 \sin 2 x}{4} + \cos 2 x + c . ▴$
Commented by bemath last updated on 12/Nov/20
$(vii) ((2x+4)/((x+1)(x−2))) = (p/(x+1)) + (q/(x−2)) → { ((p = [((2x+4)/(x−2)) ]_(x = −1) = −(2/3))),((q = [ ((2x+4)/(x+1)) ]_(x = 2) = (8/3))) :} I = ∫ (8/(3(x−2))) − (2/(3(x+1))) dx I = (8/3) ln ∣x−2∣ − (2/3)ln ∣x+1∣ + c I = (2/3)ln((((x−2)^4 )/(∣x+1∣)) ) + c . ⧫$
$(v i i) \frac{2 x + 4}{(x + 1) (x - 2)} = \frac{p}{x + 1} + \frac{q}{x - 2}$ $\to {\begin{cases} p = {[\frac{2 x + 4}{x - 2}]}_{x = - 1} = - \frac{2}{3} \\ q = {[\frac{2 x + 4}{x + 1}]}_{x = 2} = \frac{8}{3} \end{cases}$ $I = \int \frac{8}{3 (x - 2)} - \frac{2}{3 (x + 1)} d x$ $I = \frac{8}{3} \ln ∣ x - 2 ∣ - \frac{2}{3} \ln ∣ x + 1 ∣ + c$ $I = \frac{2}{3} \ln (\frac{{(x - 2)}^{4}}{∣ x + 1 ∣}) + c . ⧫$
	Answered by Dwaipayan Shikari last updated on 12/Nov/20

	$\int_{1}^{2} \frac{12 x}{{(9 - 2 x^{2})}^{2}} d x (9 - 2 x^{2} = t \Rightarrow - 4 x = \frac{d t}{d x}$ $= 3 \int_{1}^{7} \frac{d t}{t^{2}} = {[- \frac{3}{t}]}_{1}^{7}$ $= - \frac{3}{7} + 3 = \frac{18}{7}$
	Answered by mathmax by abdo last updated on 12/Nov/20
	$y=(1/(9−2x^2 )) ⇒y^′ =−((−4x)/((9−2x^2 )^2 )) =((4x)/((9−2x^2 )^2 )) ⇒ ∫_1 ^2 ((12xdx)/((9−2x^2 )^2 )) =3 ∫_1 ^2 ((4x)/((9−2x^2 )^2 ))dx =3[(1/(9−2x^2 ))]_1 ^2 =3{1−(1/7)} =3.(6/7) =((18)/7)$
	$y = \frac{1}{9 - {2 x}^{2}} \Rightarrow y^{^{'}} = - \frac{- 4 x}{{(9 - {2 x}^{2})}^{2}} = \frac{4 x}{{(9 - {2 x}^{2})}^{2}} \Rightarrow$ $\int_{1}^{2} \frac{12 xdx}{{(9 - {2 x}^{2})}^{2}} = 3 \int_{1}^{2} \frac{4 x}{{(9 - {2 x}^{2})}^{2}} dx = 3 {[\frac{1}{9 - {2 x}^{2}}]}_{1}^{2} = 3 {1 - \frac{1}{7}} = 3 . \frac{6}{7}$ $= \frac{18}{7}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum