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Q: What is the integral of tan cubed x secx dx?

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d/dx(uv)=u*dv/dx+v*du/dxd/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]-The derivative of tanx is:d/dx(tan u)=[sec(u)]2*d/dx(u)d/dx(tan x)=[sec(x)]2*d/dx(x)d/dx(tan x)=[sec(x)]2*(1)d/dx(tan x)=(sec(x))2=sec2(x)-The derivative of secx is:d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)d/dx(sec x)=[sec(x)tan(x)]*(1)d/dx(sec x)=sec(x)tan(x)d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)

Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|

XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------

integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C

Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C

Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant

convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx

In this specific example one would need to use the u substitution method. * Set u to be x - 3 * Derive x - 3 * u = x - 3 * du = dx Now that we have integrated u we can remove the x - 3 and substitute in u and remove the dx and substitute in du. This is what we have after substituting: * (the integrand of) tan(u)du Now integrate tan(u)du * the Integral of tan(u)du is: * sec2(u) Now resubstitute what we set as u. In this case we set x - 3 to u. This will give us our final answer and integral of tan(x-3)dx. * sec2(x - 3)

for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..

âˆ« tan(x) dx = -ln(cos(x)) + C C is the constant of integration.

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