How should we do this integral v/(1-v^4)?
Let I=yInteg.v/(1-v^4)Let v^2 = t. Then. 2v.dv = dtSo. v.dv = (1/2)dtSo. I = (1/2) integ.dt/(1-t^2)= (1/2).(1/2).log{(1+t)/(1-t)} +c= (1/4).log{(1+v^2)/(1-v^2) +c.
Line integral?
Evaluate the line integral by the following method. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 4) (a) directly
How do I integrate (x^4+1) / (x^6+1)?
Let [math]\displaystyle I = \int \dfrac{x^4 + 1}{x^6 + 1} \, dx[/math][math]\displaystyle = \int \dfrac{x^4 + 1 - x^2 + x^2}{x^6 + 1} \, dx[/math][math]\displaystyle = \int \dfrac{x^4 - x^2 + 1}{x^6 + 1} \, dx + \underbrace{\int \dfrac{x^2}{x^6 + 1} \, dx}_{I_1}[/math][math]\displaystyle = \int \dfrac{x^4 - x^2 + 1}{(x^2 + 1)(x^4 - x^2 + 1)} \, dx + I_1[/math][math]\displaystyle = \int \dfrac{1}{x^2 + 1} \, dx + I_1[/math][math]\displaystyle = \arctan(x) + I_1[/math]Now lets solve [math]I_1[/math],Assume [math]\displaystyle x^3 = \tan(y)[/math][math]\displaystyle \implies 3x^2 \, dx = \sec^2(y) \, dy[/math]Substitution of [math]x[/math] in [math]I_1[/math] gives us,[math]\displaystyle I_1 = \int \dfrac{\sec^2(y)}{3(\tan^2(y) + 1)} \, dy[/math][math]\displaystyle = \int \dfrac{1}{3} \, dy[/math][math]\displaystyle = \dfrac{1}{3} y[/math]So, [math]\displaystyle I_1 = \dfrac{1}{3} \arctan(x^3)[/math][math]\implies \displaystyle \bbox[#AFA]{I = \arctan(x) + \dfrac{1}{3} \arctan(x^3) + C}[/math] (where [math]C[/math] is the constant of indefinite integration)
How to integrate ((4-x^2)) ^1/2?
I will show you two methods to solve [math]\int \sqrt{4-x^2} \ dx[/math]Method 1 : Substitution.Let us first find permissible values of [math]x[/math] in [math]\sqrt{4-x^2}[/math]As we are dealing with real numbers, we cannot take square root of negative number. So, solving for [math]4-x^2\geq 0[/math] gives [math]x\in [-2,2][/math]We can replace [math]x[/math] by any function whose range is [math][-2,2][/math] and best would be replacing [math]x[/math] by [math]2 \sin \Theta [/math][math]x=2\sin \Theta \Rightarrow dx=2\cos \Theta \, d\Theta [/math]Substituting all these in above integration,[math]\int \sqrt{4-4\sin ^{2}\Theta} * 2\cos \Theta \, d\Theta [/math][math]=\int 4\cos ^2 \Theta \, d\Theta =2\int (1+\cos 2\Theta )\, d\Theta =2\Theta +\sin 2\Theta + constant[/math]Back substituting gives integral as [math]2\arcsin \frac{x}{2} + \frac{x}{2}\sqrt{4-x^2} + constant.[/math][math](\sin 2\Theta =2\sin \Theta \cos \Theta =x*\frac{\sqrt{4-x^2}}{2})[/math]Method 2 : Using Formula directly.[math]\int \sqrt{a^2-x^2}\, dx=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}\sin ^{-1}(\frac{x}{a}) + C[/math]Here, put [math]a=2[/math] to get same result as in Method 1.
How do I integrate ((x^4 - x) ^(1/4)) /x^5?
Sorry for the dirty hand writing .Please forgive me☺
Find the integral of (4-x^2)^3/2 dx?
Let x=4sin Θ , Θ=arcsin x/4 dx = 4 cos Θ dΘ (4-x^2)^3/2 = (4-4sin² Θ)^3/2 =4 (1-sin²Θ)^3/2 =4 cos³ Θ ∫(4-x^2)^3/2 dx =∫(4 cos³Θ 4 cos Θ dΘ =16 ∫ cos Θ^4 dΘ-------------(1) cos²Θ=(1/2)(1+cos 2Θ) (cos Θ)^4 = (1/4)(1 +2cos2Θ+cos²2Θ) Substitute in (1) (1)= 16 ∫(1/4)(1 +2cos2Θ+cos²2Θ) dΘ =4 ∫(1 +2cos2Θ+cos²2Θ) dΘ =4 ∫(1 +2cos2Θ+(1/2)(1+cos4Θ) dΘ =4 ∫(1 +2cos2Θ+(1/2)+(1/2)cos4Θ) dΘ =4 ∫(3/2) +2cos2Θ +(1/2)cos4Θ) dΘ =4 [ 3x/2 + sin 2Θ +(1/8)sin 4Θ] from x = -1 to 1 Use x = arcsin x/4 from x = -1 to 1 Substitute into the above equation. You will get the answer.
What is the integration of 1/ (x^1/2+x^1/4)?
[math]I=\int\dfrac{1}{x^\frac{1}{2}+x^\frac{1}{4}}dx[/math]To simplify the integral, substitute [math]u=x^\frac{1}{4}[/math]from this you get,[math]du=\frac{1}{4}x^\frac{-3}{4}dx[/math][math]dx=4x^\frac{3}{4}du=4(x^\frac{1}{4})^3du=4u^3du[/math]also,[math]u^2=x^\frac{1}{2}[/math]substitute these into the integral to get[math]I=4\int\dfrac{u^3}{u^2+u}du=4\int\dfrac{u^2}{u+1}du[/math]to simplify the denominator, use another substitution,let, [math]v=u+1[/math] and so [math]dv=du[/math]the integral becomes,[math]I=4\int\dfrac{(v-1)^2}{v}dv=4\int v+\dfrac{1}{v}-2 dv[/math][math]I=4(\dfrac{v^2}{2}+lnv-2v+C)[/math][math]=2v^2+4lnv-8v+4C[/math][math]=2(u+1)^2+4ln(u+1)-8(u+1)+C[/math][math]=2(x^\frac{1}{4}+1)^2+4ln(x^\frac{1}{4}+1)-8x^\frac{1}{4}-8+C[/math][math]-8[/math] can be combined with the constant of integration [math]C[/math] to make a new constant[math]=2(x^\frac{1}{4}+1)^2+4ln(x^\frac{1}{4}+1)-8x^\frac{1}{4}+C[/math]So, finally[math] \int\dfrac{1}{x^\frac{1}{2}+x^\frac{1}{4}}dx= \boxed{ 2(x^\frac{1}{4}+1)^2+4ln(x^\frac{1}{4}+1)-8x^\frac{1}{4}+C }[/math]