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How do you find the integral of int sin x.tan x dx?
Solution:
We will find the integral of int sin x.tan x dx
Let see, how we can solve this.
As we know that tan x is defined as
tan x = sin x/ cos x
Thus,
∫ sin x tan x dx = ∫sinx (sin x/ cos x) dx ∫sin x tan x dx=∫sin2x sec x ----------->(since 1/cos x = sec x)
∫sin x tan x dx = ∫secx (1−cos2x) =∫(sec x−cos x)dx
As we know that
∫sec x.dx = ln|tan x+sec x| and
∫cos x dx = sin x+C
Now we are substituting in the above equation and we get,
∫sin x tan x dx=ln|tan x+sec x|−sin x+C
Thus, the integral of sin x.tan x dx is ln |tan x+sec x| − sin x+C
How do you find the integral of int sin x.tan x dx?
Summary:
The integral of sin x . tan x dx= ∫sinx tanx dx=ln|tan x+sec x|−sinx+C
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