Menu Close

Question-203434

Tinku Tara 0 Comments
Pin




Question Number 203434 by sonukgindia last updated on 19/Jan/24
Answered by witcher3 last updated on 20/Jan/24
=∫((−2sin(2x)sin(3x))/(1+2cos(3x)))dx =((sin(2x))/3)ln(1+2cos(3x))−(1/6)∫sin(x)cos(x)ln(1+2cos(3x))dx ln(1+2cos(3x))=ln(1+2cos(x)(2cos^2 (x)−1)−4(1−cos^2 (x))cos(x)) =ln(1+8cos^3 (x)−6cos(x)) cos(x)=u ∫ =(1/6)∫uln(1+8u^3 −6u)du 1+8u^3 −6u=8Π_(w∣1+8w^3 −6w=0) (u−w) =(1/6)∫uln(8)+Σ_w uln(u−w)du ∫uln(u−w)=∫(u−w)ln(u−w)+wln(u−w)du =(1/2)(u−w)^2 ln(u−w)−(1/4)(u−w) ((sin(2x))/3)ln(1+2cos(3x))+(1/4)cos^2 (x)ln(2)+(1/6)Σ_(w∣8w^3 −6w+1=0) (cos(x)−w)ln(cos(x)−w)−(1/4)(cos(x)−w) w can bee expressed withe cardon
$$=\int\frac{−\mathrm{2sin}\left(\mathrm{2x}\right)\mathrm{sin}\left(\mathrm{3x}\right)}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)}\mathrm{dx} \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{3}}\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)\right)−\frac{\mathrm{1}}{\mathrm{6}}\int\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)\right)\mathrm{dx} \\ $$$$\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)\right)=\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{x}\right)\left(\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{1}\right)−\mathrm{4}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\right)\mathrm{cos}\left(\mathrm{x}\right)\right) \\ $$$$=\mathrm{ln}\left(\mathrm{1}+\mathrm{8cos}^{\mathrm{3}} \left(\mathrm{x}\right)−\mathrm{6cos}\left(\mathrm{x}\right)\right) \\ $$$$\mathrm{cos}\left(\mathrm{x}\right)=\mathrm{u} \\ $$$$\int \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int\mathrm{uln}\left(\mathrm{1}+\mathrm{8u}^{\mathrm{3}} −\mathrm{6u}\right)\mathrm{du} \\ $$$$\mathrm{1}+\mathrm{8u}^{\mathrm{3}} −\mathrm{6u}=\mathrm{8}\underset{\mathrm{w}\mid\mathrm{1}+\mathrm{8w}^{\mathrm{3}} −\mathrm{6w}=\mathrm{0}} {\prod}\left(\mathrm{u}−\mathrm{w}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int\mathrm{uln}\left(\mathrm{8}\right)+\underset{\mathrm{w}} {\sum}\mathrm{uln}\left(\mathrm{u}−\mathrm{w}\right)\mathrm{du} \\ $$$$\int\mathrm{uln}\left(\mathrm{u}−\mathrm{w}\right)=\int\left(\mathrm{u}−\mathrm{w}\right)\mathrm{ln}\left(\mathrm{u}−\mathrm{w}\right)+\mathrm{wln}\left(\mathrm{u}−\mathrm{w}\right)\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{u}−\mathrm{w}\right)^{\mathrm{2}} \mathrm{ln}\left(\mathrm{u}−\mathrm{w}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{u}−\mathrm{w}\right) \\ $$$$\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{3}}\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)\right)+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{6}}\underset{\mathrm{w}\mid\mathrm{8w}^{\mathrm{3}} −\mathrm{6w}+\mathrm{1}=\mathrm{0}} {\sum}\left(\mathrm{cos}\left(\mathrm{x}\right)−\mathrm{w}\right)\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)−\mathrm{w}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}\left(\mathrm{x}\right)−\mathrm{w}\right) \\ $$$$\mathrm{w}\:\mathrm{can}\:\mathrm{bee}\:\mathrm{expressed}\:\mathrm{withe}\:\mathrm{cardon} \\ $$
Answered by MathematicalUser2357 last updated on 30/Jan/24
((sin 2x)/3)ln(1+2cos 3x)+(1/4)cos^2 x ln 2+(1/6)Σ_(8w^3 −6w+1=0) ((cos x−w)ln(cos x−w)−(1/4)(cos x−w))+C
$$\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{3}{x}\right)+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}^{\mathrm{2}} {x}\:\mathrm{ln}\:\mathrm{2}+\frac{\mathrm{1}}{\mathrm{6}}\underset{\mathrm{8}{w}^{\mathrm{3}} −\mathrm{6}{w}+\mathrm{1}=\mathrm{0}} {\sum}\left(\left(\mathrm{cos}\:{x}−{w}\right)\mathrm{ln}\left(\mathrm{cos}\:{x}−{w}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}\:{x}−{w}\right)\right)+{C} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *