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

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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