Integrate [(1+tanx)^2]dx?
By expanding: ∫ [1 + tan(x)]^2 dx = ∫ [tan^2(x) + 2tan(x) + 1] dx = ∫ [sec^2(x) + 2tan(x)] dx, since tan^2(x) + 1 = sec^2(x) = ∫ sec^2(x) dx + 2 ∫ tan(x) dx, by splitting up the integral = tan(x) - 2ln|cos(x)| + C. So, you are correct! Also, notice that: ln|cos(x)| = ln|1/sec(x)| = -ln|sec(x)|, so: tan(x) - 2ln|cos(x)| + C = tan(x) + 2ln|sec(x)| + C, which shows that MechEng and my answer are equivalent. I hope this helps!
What is the integration of tanx^1/2?
Given:-[math]\displaystyle\int \sqrt{tanx} dx[/math]Put [math]\qquad tanx = t^2[/math]So [math]\qquad sec^2xdx = 2tdt[/math]Therefore u will get:-In part I:-Assume [math]\qquad t - \dfrac{1}{t} = p[/math]So [math]\qquad (1 + \dfrac{1}{t^2}) dt = dp[/math][math]\displaystyle\int \dfrac{dp}{p^2 + 2}[/math] [math]———(1)[/math]In part II:-Assume [math]\qquad t + \dfrac{1}{t} = q[/math]So [math]\qquad (1 - \dfrac{1}{t^2}) dt = dq[/math][math]\displaystyle\int \dfrac{dq}{q^2 - 2}[/math] [math]———(2)[/math]Integrate parts (1) and (2) and then finally put values of [math]p[/math], [math]q[/math] and [math]t[/math] respectively.U will get :-[math]\dfrac{1}{√2} tan^{-1} \Big( \dfrac{√tanx - √cotx}{√2} \Big) + \dfrac{1}{2√2} ln \Big( \dfrac{√tanx + √cotx - √2}{√tanx + √cotx + √2} \Big) + C [/math]
How do we integrate (tanx) ^1/2 + (cotx) ^1/2?
Convert them to sine and cosine, and then take LCM.0:00-0:01Thanks
How do I integrate 1/(x tanx+1) ^2?
You just require two things to know to solve this problem.Integration by substitution. Here you substitute some funtion of x with some other variable,say t, and change the integral accordingly. This is up to your intuition substituting what for what will help you integrate faster.Integration by parts, which says that if f(x) and g(x) are two continuous and differentiable functions on a certain interval I of real numbers, then ∫f(x)g(x)dx= f(x)∫g(x)dx - ∫[f’(x)×∫g(x)dx]dx, where f'(x) denotes the first order derivative of f(x).Now, keeping the above points in mind, I'm providing you a picture solution of the question:
What is integral tan x / (1 + m^2 tan^2 x) dx from 0 to pi/2?
Let [math]\displaystyle I = \int_0^{\frac{\pi}{2}} \dfrac{\tan(x)}{1 + m^2\tan^2(x)} \, dx[/math][math]\displaystyle = \int_0^{\frac{\pi}{2}} \dfrac{\frac{\sin(x)}{\cos(x)}}{1 + \frac{m^2\sin^2(x)}{\cos^2(x)}} \, dx[/math][math]\displaystyle = \int_0^{\frac{\pi}{2}} \dfrac{\sin(x)\cos(x)}{\cos^2(x) + m^2\sin^2(x)} \, dx[/math][math]\displaystyle = \int_0^{\frac{\pi}{2}} \dfrac{\sin(x)\cos(x)}{\cos^2(x) + m^2( 1 - \cos^2(x))} \, dx[/math][math]\displaystyle = \int_0^{\frac{\pi}{2}} \dfrac{\sin(x)\cos(x)}{m^2 + (1 - m^2)\cos^2(x)} \, dx[/math]Let [math]\cos(x) = y[/math][math]\implies -\sin(x) \, dx = dy[/math]At [math]x = 0[/math], [math]y = 1[/math]and [math]x = \frac{\pi}{2}, y = 0[/math]Substituting above value in [math]I[/math], we get,[math]\displaystyle I = \int_1^0 \dfrac{-y}{m^2 + (1 - m^2)y^2} \, dy[/math][math]\displaystyle = \int_0^1 \dfrac{2(1 - m^2)y}{2(1 - m^2)(m^2 + (1 - m^2)y^2)} \, dy[/math][math]\displaystyle = \int_0^1 \dfrac{1}{2(1 - m^2)(m^2 + (1 - m^2)y^2)} \, d(m^2 + (1 - m^2)y^2)[/math][math]\displaystyle = \dfrac{1}{2(1 - m^2)} \ln(m^2 + (1 - m^2)y^2) \bigg|_0^1[/math][math]\displaystyle = \dfrac{1}{2(1 - m^2)} \ln(m^2 + (1 - m^2)) - \dfrac{1}{2(1 - m^2)} \ln(m^2)[/math][math]\displaystyle = -\dfrac{1}{1 - m^2} \ln(m)[/math][math]\displaystyle \implies I = \dfrac{1}{m^2 - 1} \ln(m)[/math]
What is the integration of (tanx)^-1/3?
(tan(x))^(-1/3)=surd(cot(x),3)and the answer is1/2*ln(surd(cot(x)^2,3)+1)-1/4*ln(surd(cot(x)^4,3)-surd(cot(x)^2,3)+1)-1/2*sqrt(3)*arctan(1/3*(2*surd(cot(x)^2,3)-1)*sqrt(3))orplus arbitrary constant, of course.How to obtain —- denote y = tan(x), check Chebyshev's conditions, use appropriate substitution and integrate rational fraction function.
What is the integral of 1/(1+tan x)?
It is a simple problem.. all the integration questions just require some time and the good use of concepts..Best of luck ✌
Integral of: (tanx)^1/2 + (cotx)^1/2 dx ?
The above answer is wrong because exponents come before division by default. Integrate this expression using standard integrals: ∫ [(tanx)¹ / 2 + (cotx)¹ / 2] dx = ∫ (tanx / 2 + cotx / 2) dx ∫ [(tanx)¹ / 2 + (cotx)¹ / 2] dx = ∫ tanx dx / 2 + ∫ cotx dx / 2 ∫ [(tanx)¹ / 2 + (cotx)¹ / 2] dx = -ln|cosx| / 2 + ln|sinx| / 2 + C ∫ [(tanx)¹ / 2 + (cotx)¹ / 2] dx = ln|sinx| / 2 - ln|cosx| / 2 + C ∫ [(tanx)¹ / 2 + (cotx)¹ / 2] dx = (ln|sinx| - ln|cosx|) / 2 + C ∫ [(tanx)¹ / 2 + (cotx)¹ / 2] dx = ln|sinx / cosx| / 2 + C ∫ [(tanx)¹ / 2 + (cotx)¹ / 2] dx = ln|tanx| / 2 + C
What is the integration of tanx/1+tanx+tan^2x?
By converting whole to sinx and cosx and by further substitution we can integrate.Thank You☺