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How do you integrate #Sinx * Tanx#?

Answer 1

Evelyn Agard

The answer is #=ln (∣tanx+secx∣)-sinx +C#

We need, #secx=1/cosx#

#cos^2x+sin^2x=1#

#tanx=sinx/cosx#

#(tanx)'=sec^2x#

#(secx)'=tanx secx#

#intsinxtanxdx=int(sinx*sinxdx)/cosx#

#=intsecxsin^2xdx#

#=intsecx(1-cos^2x)dx#

#=int(secx-cosx)dx=intsecxdx-intcosxdx#

For the integral of #secx#, multiply top and bottom by #(tanx+secx)#

#intsecxdx=int(secx(tanx+secx)dx)/(tanx +secx)#

Let #u=tanx +secx#

#du=(sec^2x+secxtanx)dx#

#intsecxdx=int(du)/u#

#=lnu = ln (∣tanx+secx∣)#

And finally,

#intsinxtanxdx= ln (∣tanx+secx∣)-sinx +C#

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Answer 2

Adam Adkins

To integrate sin(x) * tan(x), you can use substitution. Let u = sin(x), then du = cos(x) dx. Therefore, the integral becomes ∫u * (du/u) = ∫du = u + C. Substituting back sin(x) for u, the result is sin(x) + C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.