Integral sqrt(x^2-4) dx?
What Aj did was completely wrong, learn the formulas correctly its (x^2-4)^1/2 , the WHOLE thing is to the power, that rule only works when its something like 3x^3 (NO BRACKETS, a single term ) one a similar note, there is a formula for integrating linear functions to a power, that is integral (ax+b)^n= {(ax+b)^(n+1)}/ a(n+1), however this formula only works for linear functions(no x^2, x^3, x^1/2 etc)
Integral of sqrt(x^2-4)/x?
∫√(x²-4)/x.dx=∫√4(x²/4-1)/x.dx =2∫√{(x/2)²-1)/x.dx Let x/2= secz →dx/2 =secz tanzdz or dx =2secztanzdz Substituting we get ∫√(x²-4)/x.dx =2∫{√(sec²z-1)/(2secz)}2secztanzdz =2∫{√(tan²z)/(2secz)}2secztanzdz =2∫{(tanz)tanzdz =2∫tan²zdz =2∫(sec²z−1)dz =2∫(sec²z)dz−2∫dz =2tanz−2z=2(tanz−z) Since x/2 = secz Hence tanz =√(sec²z-1)=√{(x/2)²-1} =(1/2)√{x²-4} and z=sec⁻¹(x/2) Hence ∫√(x²-4)/x.dx=2(tanz−z)+C =2[{(1/2)√(x²-4)}−sec⁻¹(x/2)]+C
What is the integral of x^2/sqrt (x^2 + 1)?
I will help you reduce this in standard form.[math]I=\dfrac{x^2+1-1}{\sqrt{x^2+1}}[/math][math]=> I=\sqrt{x^2+1}-\dfrac{1}{\sqrt{x^2+1}}[/math]Now take this forward from here.
What is the integral of: dx/sqrt(25x^2-4), where x>2/5?
∫dx /√(25x^2 - 4) let 5x = 2sec u ==> sec u = (5x/2) and tan u = (1/2)√(25x^2 - 4) 5dx = 2sec(u) tan(u) du dx = (2/5) sec(u) tan(u) du 2/5∫sec(u) tan(u) du /√(4sec^2(u) - 4) = 2/5∫sec(u) tan(u) du /2√(sec^2(u) - 1) = 1/5∫sec(u) tan(u) du /tan (u) = 1/5∫sec(u) du = (1/5) ln I sec u + tan u I + C substitute sec u = (5x/2) and tan u = (1/2)√(25x^2 - 4) (1/5) ln I 1/2 [ 5x + √(25x^2 - 4) I + C
Problem with solving integral (sqrt(x/(4-x)) dx?
sqrt(x / (4 - x)) * dx sqrt(x) * dx / sqrt(4 - x) sqrt(4 - x) = u 4 - x = u^2 4 - u^2 = x -2u * du = dx sqrt(4 - u^2) * (-2u) * du / u -2 * sqrt(4 - u^2) * du u = 2 * sin(t) du = 2 * cos(t) * dt -2 * sqrt(4 - 4 * sin(t)^2) * 2 * cos(t) * dt => -4 * sqrt(4 * cos(t)^2) * cos(t) * dt => -4 * 2 * cos(t) * cos(t) * dt => -8 * cos(t)^2 * dt => -8 * (1/2) * (1 + cos(2t)) * dt => -4 * (1 + cos(2t)) * dt => -4 * dt - 4 * cos(2t) * dt Integrate -4t - 4 * (1/2) * sin(2t) + C= > -4t - 4 * sin(t) * cos(t) + C u = 2 * sin(t) u/2 = sin(t) (u/2)^2 = sin(t)^2 (u/2)^2 = 1 - cos(t)^2 cos(t)^2 = (4 - u^2) / 4 cos(t) = sqrt(4 - u^2) / 2 -4t - 4 * sin(t) * cos(t) + C -4 * arcsin(u/2) - 4 * (1/2) * (1/2) * u * sqrt(4 - u^2) + C => -4 * arcsin(u/2) - u * sqrt(4 - u^2) + C u = sqrt(4 - x) -4 * arcsin(sqrt(4 - x) / 2) - sqrt(4 - x) * sqrt(4 - (4 - x)) + C => -4 * arcsin(sqrt(4 - x) / 2) - sqrt(x * (4 - x)) + C I got the same thing you did. If you can take the derivative of your answer and get the original function, then you're fine.
How do I integrate (dx/ (x^2+4) (sqrt (4x^2+1)))?
This is a standard question of the form 1/ quadratic multiplied by square root quadratic.The substitution is x = 1/t and after that one more substitution is required , the quantity in square root must be substituted as u ^2.And the question gets solved.
Integrate (sqrt x )/(lnx) and 1/ (sqrt 49 + x^2) and (sin^5x)(cos^18x)?
1. Taking u = ln x and dv = √x, and integrating by parts, ...∫ ( ln x ).√x dx = [ ( ln x ).( 2/3 )* x.√x ] - ∫ [ ( 2/3 )* x√x * ( 1/x ) ] dx = ( 2/3 )* x.√x. ln x - ( 2/3 )* ∫ √x dx = ( 2/3 )* x.√x. ln x - ( 2/3 )* x.√x + C = ( 2/3 )* x.√x [ ( ln x ) - ( 2/3 ) ] + C. .....Ans. ......................................... 2. ∫ 1 / √( a² + u² ) du = ln [ u + √( a² + u² ) ] ...∫ 1 / √ ( 7² + x² ) dx = ln [ x + √( 49 + x² ) ] + C. ....Alternatively, you can substitute x = 7* tan θ so that dx = 7* sec² θ dθ. ......................................... 3. I = ∫ ( sin^5 x )( cos^18 x ) dx ......= ∫ ( sin^4 x )( cos^18 x ). sin x dx ......= ∫ ( s² )². c^18. sin x dx ......= ∫ ( 1 - c² )². c^18. sin x dx ......= ∫ ( 1 - 2 c² + c^4 ). c^18. sin x dx....(1) ......................................... Putting u = cos x gives du = - sin x dx, i.e., sin x dx = - du. ......................................... From (1), I = ∫ ( 1 - 2u² + u^4 ).u^18 ( - du ) = ∫ ( 2u^2 - u^4 - 1 ). u^18 du = ∫ ( 2 u^20 - u^22 - u^18 ) du = 2 ( u^21 / 21 ) - ( u^23 / 23 ) - ( u^19 / 19 ) + C = ( 2/21 ) cos^21 x - ( 1/23 ) cos^23 x - ( 1/19 ) cos^19 x + C. ......................................... Happy To Help ! .........................................