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# integrate cos2x/(1+tanx) w.r.t x 10 years ago

Hi Suchita,

Here write tanx = six/cosx, and you have the integral

cos3x/(1+sinx).

Make the substitution sinx = t. You hae cosxdx = dt.

So integral reduces to (1-t2)/(1-t) = 1+t.

Now integral of that is t + t2/2 + constant, where t = sinx.

All the best,

Regards,

10 years ago

Hi Guys,

That was a very very silly mistake on my part.

So now the integral is cos3x/(sinx+cosx).

Multiply Nr and Dr by sinx+cosx, and you have {(cos3xsinx+cos4x) / (1+sin2x)}--------- (As [sinx+cosx]2=1+sin2x)

The Nr is (cos2xsin2x)/2 + (1+cos2x)2/4 = (1/4)(1+cos2x)sin2x + (1/4)(1+cos22x+2cos2x).

So we split this into 5 integrals:

I1 = sin2x/(1+sin2x) = 1- (1/{1+sin2x})----------- standard integral of 1/(asinx+bcosx) form.

I2 = sin2x*cos2x/(1+sin2x) ------------ make the substitution 1+sin2x=t (and we get a simple integrable function)

I3 = 1/(1+sin2x) ---------- standard integral same as in I1. (But note that it would get cancelled by that in I1, because of the + and - on the two integrals. So actually no need to integrate this).

I4 = cos22x/(1+sin2x) = (1-sin22x)/(1+sin2x) = 1-sin2x ------------------ (integrate directly by formula for sinx)

I5 = cos2x/(1+sin2x) ---------- (substitute sin2x = t, and integrate directly).

All though it looks a bit lengthy, it actually is quite simple. We actually only integrate I2, I4, and I5 all of which are very easily integrable by the method mentioned above. And please try avoiding silly mistakes as I'd done previously.

Hope that helps.

All the best.

Regards,